SOLUTIONS OF SELECTED EXERCISES IN T. TAO’S NONLINEAR
DISPERSIVE EQUATIONS
ARICK SHAO
This is a list of solutions to some of the exercises in the book Nonlinear Dispersive
Equations: Local and Global Analysis, by T. Tao. 1 Many of the later problems (beginning
from Section 2.3) were done in collaboration with the Nonlinear Dispersive Equations
reading group (Jordan Bell, David Reiss, Kyle Thompson) at the University of Toronto.
Chapter 1: Ordinary Differential Equations
1.2. First of all, fixed points are unique, since if u, v ∈ X are fixed points of Φ, then
d(u, v) = d(Φ(u), Φ(v)) ≤ cd(u, v),
which is only possible when d(u, v) = 0, i.e., u = v.
Next, fix any u0 ∈ X, and define recursively uk+1 = Φuk . By induction and the contraction mapping property, we have d(uk , uk+1 ) ≤ ck d(u0 , u1 ), and hence for any m ≤ n,
d(um , un ) ≤
n−1
X
d(uk , uk+1 ) ≤ d(u0 , u1 )
k=m
n−1
X
ck ≤
k=m
cm d(u0 , u1 )
.
1−c
In particular, {uk } is a Cauchy sequence, so there is some u ∈ X such that uk → u. Since
contraction mappings are clearly continuous (by the contraction property), then
Φ(u) = lim Φ(uk ) = lim uk+1 = u,
k
k
and hence u is the fixed point of Φ.
Finally, for any v ∈ X, we define v0 = v and vk+1 = Φvk , as before. By continuity,
d(v, u) = lim d(v0 , vk ) ≤ lim
k
k
k−1
X
d(vi , vi+1 ) ≤ d(v0 , v1 )
X
ci =
i
i=0
1
d(v, Φ(v)).
1−c
1.3. Let A = ∇Φ(x0 ). Since A is nonsingular by assumption, there exists λ > 0 such that
2λkA−1 k ≤ 1, with k · k denoting the operator norm.
Fix y ∈ D, and define the map ϕy : D → D by
2
ϕy (x) = x + A−1 [y − Φ(x)].
Note that x is a fixed point of ϕy if and only if Φ(x) = y. Taking the differential, we obtain
∇ϕy (x) = I − A−1 ∇Φ(x) = A−1 [A − ∇Φ(x)].
By continuity, there exists a neighborhood U of x0 such that kA − ∇Φ(x)k < λ for all x ∈ U.
Consequently, for any x ∈ U, we have the bound
1
k∇ϕy (x)k ≤ kA−1 kkA − ∇Φ(x)k < .
2
1
See [5].
2The solution was obtained mostly from [3].
1
2
ARICK SHAO
It follows that ϕy is Lipschitz on U, with Lipschitz constant less than 1/2, i.e.,
|ϕy (x1 ) − ϕy (x2 )| ≤
1
|x1 − x2 |,
2
x1 , x2 ∈ U.
In particular, if Φ(x1 ) = Φ(x2 ), then ϕy (x1 )−ϕy (x2 ) = x1 −x2 , and hence 2|x1 −x2 | ≤ |x1 −x2 |,
i.e., x1 = x2 . From this, we conclude that Φ is one-to-one on U.
Next, fix any x1 ∈ U, let V = Φ(U), and let y1 = Φ(x1 ) ∈ V. Choose r > 0 such that
B̄ = B(x1 , r) is contained in U. 3 If y ∈ D is such that |y − y1 | < λr, then for any x ∈ B̄,
|ϕy (x) − x1 | ≤ |ϕy (x) − ϕy (x1 )| + |ϕy (x1 ) − x1 | ≤
1
|x − x0 | + kA−1 k|y − y0 | ≤ r.
2
Therefore, ϕy maps from B̄ into B̄. In particular, ϕy | B̄ is a contraction mapping on the
complete metric space B̄, so that ϕy has a unique fixed point z ∈ B̄. This implies that
Φ(z) = y, so that y ∈ V. With this, we have now proved that V is open, and that Φ is a
one-to-one mapping from U onto V.
Let Ψ : V → U be the inverse of Φ. Let y, y + k ∈ V, and define
x = Ψ(y),
x + h = Ψ(y + k).
Since
|h − A−1 k| = |h + A−1 [Φ(x) − Φ(x + h)]| = |ϕy (x + h) − ϕy (x)| ≤
1
|h|,
2
it follows that |h| ≤ 2|A−1 k| ≤ λ−1 |k|. Moreover, since
kI − A−1 ∇Φ(x)k ≤ kA−1 [A − ∇Φ(x)]k ≤
1
<1
2
by all our previous assumptions, then A−1 ∇Φ(x), and hence ∇Φ(x), is invertible.
Let S = ∇Φ(x), and let T = S −1 . A direct computation yields
|Ψ(y + k) − Ψ(y) − T k| | − T [Φ(x + h) − Φ(x) − S h]|
=
|k|
|k|
kT k |Φ(x + h) − Φ(x) − S h|
≤
·
.
λ
|h|
The right-hand side goes to zero as |h| & 0. Since we have proved this for arbitrary y and
y + k ∈ V, then Ψ is differentiable on V, and ∇Ψ(y) = [∇Φ(Ψ(y))]−1 . Since both Ψ and ∇Φ
are continuous, the above formula implies that ∇Ψ is continuous, so that Ψ = Φ−1 is C 1 .
1.4. We begin by generating a solution u ∈ C 1 (I → D) using Theorem 1.7, and we proceed
by induction. Suppose u is C l for l ≤ k. Then, ∂t u = F ◦ u is C l as well, which implies that
u is C l+1 . This iterative process continues until l = k, so that u is C k+1 . By definition, then
k+1
u ∈ Cloc
(I → D), and the map S t0 (t) is k times differentiable.
1.5. The Picard existence theorem generalizes directly to higher-order quasilinear ODE,
since these can be reformulated equivalently as first-order systems. Similarly, the Picard
existence theorem (and also the Cauchy-Kowalevski theorem) extends to non-autonomous
systems, since these can be equivalently formulated as autonomous systems.
3We let B(x , r) denote the open ball of radius r about x .
0
0
SOLUTIONS
3
1.6. The infinite iteration scheme described in the problem statement can collapse, since
the time intervals ∆ti = ti − ti−1 , i ≥ 1, in each iteration step can become arbitrarily small,
depending on the growth of the solution at each step. For example, if
∞
X
∆ti < ∞,
i=1
then we have only a finite-time solution.
In particular, in the case of (1.6), we have |u(t)| = 1/(1 − t). Suppose we solve using
Picard iteration beginning at time 0 ≤ t0 < 1. Let Ω, as in the statement of Theorem 1.7,
be the ball B = B(0, 2/(1 − t0 )). In this case, F is given by F(u) = u2 , so that
kFkC 0 (B) ≤
4
,
(1 − t0 )2
kFkĊ 0,1 (B) ≤
4
.
1 − t0
(For the latter inequality, we have |F(u) − F(v)| ≤ |u + v||u − v| ≤ (|u| + |v|)|u − v|.) As a
result, we only have local existence on a time interval T min((1 − t0 )2 , (1 − t0 )) . 1 − t0 .
In particular, t0 + T will always be smaller than 1, no matter the choice of t0 , resulting in
the qualitative description of the preceding paragraph.
1.7. Assume u(t0 ) ≤ v(t0 ), and define the function
f (t) = [max(0, u(t) − v(t))]2
on I. Then, f is differentiable a.e., and when exists,
0
∂t f (t) =
2(u(t) − v(t))(u0 (t) − v0 (t))
v(t) ≥ u(t),
v(t) ≤ u(t).
Since I is compact and F is Lipschitz, then when v(t) ≤ u(t), we have
|∂t f (t)| ≤ 2|u(t) − v(t)||F(t, u(t)) − F(t, v(t))| . |u(t) − v(t)|2 .
This implies that |∂t f (t)| . | f (t)| almost everywhere on I, so by Gronwall’s inequality, then
f (t) ≤ f (t0 ) exp[(t − t0 )C]
for all t ∈ I and for some constant C > 0. Since f (t0 ) = 0 by definition, then f (t) = 0 for
all t ∈ I. In other words, u(t) ≤ v(t) for all t ∈ I.
Now, assume u(t0 ) < v(t0 ), let > 0 be a small constant such that u(t0 ) + ≤ v(t0 ), and
define g(t) = [max(0, u(t) + − v(t))]2 on I. Again, by differentiating g, we obtain
v(t) ≥ u(t) + ,
0
∂t g(t) =
2(u(t) + − v(t))(u0 (t) − v0 (t)) v(t) ≤ u(t) + .
Again recalling the Lipschitz property of F, we obtain for almost every t ∈ I that
|∂t g(t)| . |u(t) + − v(t)||F(t, u(t)) − F(t, v(t))|
. |u(t) + − v(t)||u(t) − v(t)|
. |u(t) + − v(t)|2 + |u(t) − − v(t)|
. 2 + |u(t) + − v(t)|2 ,
whenever u(t) + ≥ v(t). Thus, for some constant C > 0, independent of , we have
∂t g(t) ≤ Cg(t) + C 2 ,
∂t [e−C(t−t0 ) g(t)] ≤ Ce−C(t−t0 ) 2 ,
4
ARICK SHAO
for almost every t ∈ I. Integrating the above yields the inequalities 4
Z t
−C(t−t0 )
2
e
g(t) ≤ g(t0 ) + Ce−C(s−t0 ) ds = 2 (1 − e−C(t−t0 ) ),
t0
g(t) ≤ 2 (eC(t1 −t0 ) − 1).
Given t ∈ I, if u(t) + ≤ v(t), then u(t) < v(t) as desired, so there is nothing left to prove.
On the other hand, if u(t) + > v(t), then the definition of g and the above yield that
[u(t) + − v(t)]2 = g(t)2 ≤ 2 [eC(t1 −t0 ) − 1].
By choosing which is small with respect to C(t1 − t0 ), 5 then the above implies
u(t) + − v(t) ≤ ,
u(t) < u(t) + ≤ v(t).
2
2
1.8. With the notations of Theorem 1.10, let [t0 , t1 ] = [0, 1], let A = 2, and define
B(t) = −1,
u(t) = 1,
0 ≤ t ≤ 1.
It is clear from computation that
u(t) = 1 ≤ 2 − t = A +
Z
t
B(s)u(s)ds,
0 ≤ t ≤ 1.
t0
However, we have that
t
Z
A exp
!
B(s)ds = 2e−t ,
t0
which is smaller than u(t) = 1 for t sufficiently close to 1.
To reconcile this with Theorem 1.12, we observe that within the proof of Theorem 1.10,
we obtain the inequality
!
!
Z t
Z t
d
A+
B(s)u(s)ds ≤ B(t) A +
B(s)u(s) ,
dt
t0
t0
which reduces precisely to Theorem 1.12. However, if B is allowed to be negative, then
Z t
A+
B(s)u(s)ds
t0
needs no longer be nonnegative, which violates the hypotheses of Theorem 1.12.
1.10. First, we apply the usual Picard theory to obtain a solution u : (T − , T + ) → D to
(1.7) for which the interval of existence is maximal. We wish to show that T + = ∞ and
T − = −∞. We need only show the former, since the latter then follows from inverting the
time variable. 6 Suppose now that T + < ∞; differentiating ku(t)k2 , we obtain
∂t (1 + ku(t)k2 ) = h∂t u(t), u(t)i . kF(u(t))kku(t)k . 1 + ku(t)k2 ,
where in the last step, we applied the linear bound for F, along with Cauchy’s inequality.
Applying the differential Gronwall inequality, then we have
(1 + ku(t)k2 ) . e(t−t0 )C (1 + ku0 k2 ) ≤ e(T+ −t0 )C (1 + ku0 k2 ),
t0 ≤ t ≤ T + ,
for some constant C > 0. This contradicts Theorem 1.17, so that T + = ∞.
Since solutions to (1.7) are unique due to Theorem 1.14, and since solutions have the
time translation invariance property (due to (1.7) being an autonomous system), then the
4Note that here, we have implicitly derived a slightly more general form of the Gronwall inequality.
5Recall that C and t − t are both independent of .
1
0
6Defining v(t) = u(−t), then ∂ v(t) = −F(v(t)), and −F satisfies the same bounds hypothesized for F.
t
SOLUTIONS
5
solution maps obey the desired time translation invariance S t0 (t) = S 0 (t − t0 ). 7 That
S 0 (0) = id follows immediately by the definition of the solution map. By the uniqueness
of solutions (Theorem 1.14) and the above time translation invariance,
S 0 (t0 )S 0 (t) = S t (t0 − t)S 0 (t) = S 0 (t0 ).8
Finally, we observe that for any t, t0 ∈ R, say with t < t0 , we have the bound
"
#
Z t0
Z t0
ku(t0 ) − u(t)k ≤
kF(u(t))k .
(1 + ku(s)k)ds ≤ |t0 − t| 1 + sup ku(s)k .
t
t0 ≤s≤t
t
It follows that the solution map is locally Lipschitz.
1.11. Suppose the solution curve u : (T − , T + ) → D be maximal. Recall that the Picard
theory implies |u(t)| → ∞ as t % T + . Thus, when t nears T + , and hence u(t) is large, we
have that |F(u(t))| . |u(t)| p , and therefore
∂t |u(t)|2 ≤ hF(u(t)), u(t)i . |u(t)| p+1 ,
−p−1
2
∂t |u(t)|1−p . [|u(t)|2 ] 2 ∂t |u(t)|2 . 1.
1− p
Integrating the above, we obtain
Z T+
|u(t)|1−p = |u(t)|1−p − lim |u(s)|1−p . p
ds = T + − t.
s%T +
t
Taking the above to the 1/(1 − p) power yields our desired lower bound for near T + . The
analogous lower bound near T − can be proved similarly.
To see that this blowup rate is sharp, consider the case D = R and the nonlinearity
F(u) = (p − 1)−1 |u| p−1 u.
Consider this particular ODE, with initial condition u(0) = 1. A simple change of variables
yields the relation ∂t |u|1−p ≡ −1, hence it has the explicit solution
|u|1−p − 1 = −t,
−1
u(t) = (1 − t) p−1 .
Since u blows up at time T + = 1, this is precisely the proved blowup rate.
Furthermore, if we take instead the initial condition u(0) = −1, then this has the explicit
solution |u|1−p = (−1 − t). This blows up at T − = −1 and has proved blowup rate.
1.12. Let g(t) = log(3 + |u(t)|2 ). Differentiating this yields the inequality
∂t g(t) = (3 + |u(t)|2 )−1 hF(u(t)), u(t)i
. (3 + |u(t)|2 )−1 |u(t)|(1 + |u(t)|) log(2 + |u(t)|)
|u(t)| + |u(t)|2
· log(3 + |u(t)|2 )
3 + |u(t)|2
. g(t).
.
The differential Gronwall inequality implies that g(t) ≤ g(t0 )eC(t−t0 ) for some C > 0. Taking
the exponential of both sides of the above inequality
C(t−t0 )
3 + |u(t)|2 ≤ (3 + |u(t0 )|2 )e
.
7Both sides of the equality represent solving (1.7) forward for time t − t with the same initial data.
0
8The right-hand side represents solving (1.7) forward for time t, while the middle expression represents
solving (1.7) forward for time t, and then solving forward again with this new data by time t0 − t.
6
ARICK SHAO
In particular, this implies the growth bound
|u(t)| . exp{B exp[C(t − t0 )]},
where B and C are some positive constants, with B depending on u(t0 ). As a result, by
Theorem 1.17, solutions to this ODE exist for all times.
To that see this bound is sharp, consider the equation
∂t u = (1 + u) log(1 + u),
u(0) = e − 1,
which has explicit solution 1 + u(t) = exp exp t. In particular, this solution satisfies
|u(t)| & exp exp(t − 0).
1.13. First, by a direct calculation using the chain rule, we have
∂t [H(u(t))] = h∂t u(t), dH(u(t))i = hF(u(t)), dH(u(t))i = G(u(t))H(u(t)).
As a result, considering the nonnegative function f = H ◦ u, we have
∂t f (t) = (G ◦ u)(t) f (t).
Since G ◦ u is continuous, and since f (t) vanishes at some t0 ∈ I, then the differential
Gronwall inequality, applied at base point t0 , shows that f = H ◦ u vanishes everywhere.
Geometrically, if we interpret F as being a vector field describing the evolution of u,
then the statement has the following interpretation: if the solution curve u(t) begins on the
level set H = 0, and if hF, dHi vanishes to first order in H on the level set H = 0, with the
ratio G being continuous, 9 then u(t) remains everywhere on the level set H = 0.
1.21. We perform a standard bootstrap argument. Define the set
A = {t ∈ I|u(t) ≤ 2A}.
Since t0 ∈ A, then A is nonempty. Moreover, since u is continuous, then A is closed.
Next, suppose t ∈ A, let M be the maximum value of F on the closed unit ball of radius
2A, and let ε = A/2M. By our assumptions, we have the estimate
3
A.
2
Since u is continuous, then A is open. Since I is connected, and since A is nonempty,
closed, and open, then A = I, and hence u(t) ≤ 2A for all t ∈ I.
For counterexamples, suppose first that ε is not small. Let A = 1, let F(v) = v, and take,
for instance, ε = 1. Then, the assumed inequality is u(t) ≤ 1 + u(t), which trivially holds.
Thus, u can be any positive continuous function on I, and we have no uniform bound for u.
Next, we consider the case in which u is not continuous. Let A = 1, and let F(v) = v2 ,
so the assumed inequality is u(t) ≤ 1 + ε[u(t)]2 . Note that v ≤ 1 + v2 holds for v ≤ 1
and for sufficiently large v with respect to ε, say v ≥ Cε > 2. Thus, we can construct the
discontinuous function u by fixing t0 < t1 ∈ I and defining
t < t1
1
u(t) =
C t ≥ t1 .
u(t) ≤ A + εF(u(t)) ≤ A + εM ≤
This function clearly satisfies u(t0 ) ≤ 2A = 2 and the assumed inequality u(t) ≤ 1+ε[u(t)]2 ,
but it is also not uniformly bounded by 2 = 2A.
9In particular, F is tangent to the level set H = 0, since dH as a vector field is normal to the level sets of H.
SOLUTIONS
7
1.22. From Young’s inequality, we have the bound
θ
Bu(t)θ = [2θ B][2−θ u(t)θ ] ≤ (1 − θ)2 1−θ B 1−θ + θ2−1 u(t).
1
Plugging this into our assumed inequality for u, then we have
θ
1
u(t) ≤ 2[1 − θ2−1 ]u(t) ≤ 2A + 2(1 − θ)2 1−θ B 1−θ + 2εF(u(t)).
The desired result now follows from the above and from Exercise (1.21).
1.23. Consider an open ball U = B(u0 , δ) about u0 such that F is continuous on Ū. Then,
there is a sequence {Fm : Ū → R} of Lipschitz continuous functions such that Fm → F
uniformly. 10 Since Fm → F uniformly, the Fm ’s are uniformly bounded; in particular,
there is some M > 0 such that if u ∈ Ū, then |Fm (u)| ≤ M for all m.
We can now solve using the usual Picard theory for maximal solutions
um : (T −,m , T +,m ) → D,
∂t um (t) = Fm (um (t)),
um (t0 ) = u0 .
The next goal is to show uniform control for the T −,m ’s, the T +,m ’s, and the um ’s.
Fix now a single m, and define the constant
εm = min((2M)−1 δ, T +,m − t0 , t0 − T −,M ).
For our bootstrap argument, we define
Am = {d ∈ [0, εm ) | |um (t0 + s) − u0 | ≤ δ for all s ∈ [−d, d]}.
By definition, 0 ∈ Am , and Am is closed. Furthermore if d ∈ Am , then
Z t0 ±d
1
|um (t0 ± d) − u0 | ≤
|Fm (um (s))|ds ≤ εm M ≤ δ,
2
t0
and by continuity, it follows that a neighborhood of d in [0, εm ) is contained in Am . Therefore, Am is open, so by connectedness, then Am = I. Since the above holds for any arbitrary
m, then we have shown that um (t) ∈ Ū whenever |t − t0 | < εm .
Combining the above argument with Theorem 1.17, we see that
εm = (2M)−1 δ = ε,
[t0 − ε, t0 + ε] ⊆ (T −,m , T +,m )
11
for every m. This establishes the desired uniform bounds on the T −,m ’s and T +,m ’s. The
preceding bootstrap argument also yields uniform bounds for all the um ’s on [t0 − ε, t0 + ε].
Furthermore, we have the uniform bounds
|∂t um (t)| ≤ |Fm (um (t))| ≤ M,
|t − t0 | ≤ ε,
so the um ’s are uniformly Lipschitz and hence equicontinuous on this interval. By the
Arzela-Ascoli theorem and the above bootstrap bound, restricting to a subsequence, then
the um ’s converge uniformly to a continuous function
u : [t0 − ε, t0 + ε] → D,
ku − u0 k∞ ≤
δ
.
2
Since um (t) → u(t) and um (t0 ) → u(t0 ), then we have
Z t
Z t
Z t
u(t) = u(t0 ) +
F(u(s))ds + lim
[Fm (um (s)) − F(u(s))] = u(t0 ) +
F(u(s))ds.
t0
m
t0
t0
10For example, this can be easily proved using the Stone-Weierstrass theorem.
11If, say, T
−1
+,m − t0 ≤ (2M) δ, then the bootstrap bound implies that u(t) is uniformly bounded for all
t ∈ [t0 , T +,m ), which by Theorem 1.17 contradicts the maximality of T +,m .
8
ARICK SHAO
In particular, the limit of the integral vanishes by the dominated convergence theorem,
since we have the crude bound |Fm (um (s)) − F(u(s))| ≤ 2M. Finally, by Lemma 1.3, then u
is a classical solution to the original ODE.
1.27. Correction: We also require the following orthogonality condition for J:
hJu, Jvi = hu, vi,
12
u, v ∈ D.
Without this condition, ω can fail to be antisymmetric. For example, if
D = R2 ,
e1 = (1, 0),
then we can define Je1 = 2e2 and Je2 =
− 12 e1 ,
e2 = (0, 1),
so that
1
1
ω(e1 , e2 ) = − he1 , e1 i = − ,
ω(e2 , e1 ) = 2he2 , e2 i = 2.
2
2
With the above condition, first note that ω is indeed antisymmetric, since
ω(v, u) = hJu, vi = hJ 2 u, Jvi = −hu, Jvi = −ω(u, v).
To show bilinearity, we need only show linearity in the first variable:
13
ω(au1 + bu2 , v) = hau1 + bu2 , Jvi = ahu1 , Jvi + bhu2 , Jvi = aω(u1 , v) + bω(u2 , v).
For nondegeneracy, given u ∈ D \ {0}, then hu, ui > 0, so that ω(u, Ju) = −hu, ui , 0. As a
result, ω is a symplectic form.
Finally, to see that ∇ω H = J∇H, 14 we note that for any v ∈ D,
ω(J∇H, v) = hJ∇H, Jvi = h∇H, vi.
1.28. We induct on the dimension n of D. First of all, the desired statement is trivial for
dimension n = 0. Fix now n > 0, and suppose the conclusion is true for any dimension
strictly less than n. Then, we need only prove the same conclusion for dimension n.
Let u ∈ D \ {0}. Since ω is nondegenerate, there is some v ∈ D such that ω(u, v) = 1. 15
In addition, the restriction ω0 of ω to the subspace D0 = span{u, v} is du ∧ dv, i.e.,
ω(a1 u + b1 v, a2 u + b2 v) = a1 b2 − a2 b1 .
Consider next the symplectic complement
D0 = {w ∈ D|ω(u, w) = ω(v, w) = 0}.
Since ω is nondegenerate, then the linear functionals ωu = ω(u, ·) and ωv = ω(v, ·) map
onto R, so that their nullspaces satisfy
dim N(ωu ) = dim N(ωv ) = n − 1.
Furthermore, since ωu−v is nontrivial as well, then N(ωu ) and N(ωv ) cannot completely
coincide, and it follows that
dim D0 = dim[N(ωu ) ∩ N(ωv )] = n − 2.
By the induction hypothesis, then the restriction ω0 to D0 has the desired standard decomposition in some coordinates p j and q j , where 1 ≤ j ≤ n/2 − 1. In other words,
X
ω0 =
(dq j ∧ d p j ).
1≤i≤ n−2
2
12This may already have been covered by the condition that J is an endomorphism of D.
13Linearity in the second variable follows immediately from the antisymmetry of ω.
14Although the book asserts ∇ H = −J∇H, the minus sign should not be present here.
ω
15In particular, n ≥ 2.
SOLUTIONS
9
In particular, both D0 and D have even dimension. Since D0 and D0 are by definition
ω-orthogonal, then ω in fact also has this form:
X
ω=
(dq j ∧ d p j ) + du ∧ dv.
1≤i≤ n−2
2
1.30. Suppose u satisfies a Hamiltonian equation, with Hamiltonian H. We make the
change of variables v(t) = u(−t). By the chain rule,
∂t v(t) = −∇ω H(u(−t)) = −∇ω H(v(t)),
and hence v also satisfies a Hamiltonian equation, with Hamiltonian −H.
1.31. We can define a natural product h·, ·iD×D0 and symplectic form ω ⊕ ω0 on D × D0 by
h(u, u0 ), (v, v0 )iD×D0 = hu, viD + hu0 , v0 iD0 ,
(ω ⊕ ω0 )((u, u0 ), (v, v0 )) = ω(u, v) + ω0 (u0 , v0 ).
Define the Hamiltonion H ⊕ H 0 ∈ C 2 (D × D0 → R) by
(H ⊕ H 0 )(u, u0 ) = H(u) + H 0 (u0 ).
A standard calculation yields that its differential is
d(H ⊕ H 0 )(u, u0 ) = (dH(u), dH 0 (u0 )) ∈ D × D0 ,
so that
hd(H ⊕ H 0 )(u, u0 ), (v, v0 )iD×D0 = hdH(u), viD + hdH 0 (u0 ), v0 iD0
= ω(∇ω H(u), v) + ω0 (∇ω0 H 0 (u0 ), v0 )
= (ω ⊕ ω0 )((∇ω H(u), ∇ω0 H 0 (u0 )), (v, v0 )).
As a result,
∇ω⊕ω0 (H ⊕ H 0 )(u, u0 ) = (∇ω H(u), ∇ω0 H 0 (u0 )),
and hence u and u0 , as given in the problem, satisfy
∂t (u(t), u0 (t)) = (∇ω H(u(t)), ∇ω0 H 0 (u0 (t))) = ∇ω⊕ω0 (H ⊕ H 0 )(u(t), u0 (t)).
1.32. Let dim D = 2n, and let pi , qi , where 1 ≤ i ≤ n, denote the standard coordinates for
the symplectic space (D, ω); see Example (1.27) and Exercise (1.28).
First, suppose u ∈ C 2 (R × R × R → D), such that u(·, x, y) is a solution curve for the
Hamiltonian equation for H for every x, y ∈ R, i.e., that
∂t u(t, x, y) = ∇ω H(u(t, x, y)),
t, x, y ∈ R.
By a direct computation, we have
∂t [ω(∂ x u(t, x, y), ∂y u(t, x, y))] = ω(∂ x ∂t u(t, x, y), ∂y u(t, x, y)) + ω(∂ x u(t, x, y), ∂y ∂t u(t, x, y))
= ω(∂ x ∇ω H(u(t, x, y)), ∂y u(t, x, y))
+ ω(∂ x u(t, x, y), ∂y ∇ω H(u(t, x, y))),
where in the first equality, one can justify the Leibniz rule for ∂t by expanding ω in terms
of the qi ’s and pi ’s. By the bilinearity properties of ω, along with the definition of ∇ω , then
∂t [ω(∂ x u(t, x, y), ∂y u(t, x, y))] = h∂ x [dH(u(t, x, y))], ∂y u(t, x, y)i
− h∂ x u(t, x, y), ∂y [dH(u(t, x, y))]i
= h∇ H(u(t, x, y))[∂ x u(t, x, y)], ∂y u(t, x, y)i
2
− h∂ x u(t, x, y), ∇2 H(u(t, x, y))[∂y u(t, x, y)]i,
10
ARICK SHAO
where in the last step, we simply applied the chain rule. Treating the Hessian ∇2 H of H at
u(t, x, y) as a bilinear map, then the above becomes
∂t [ω(∂ x u(t, x, y), ∂y u(t, x, y))] = ∇2 H(u(t, x, y))[∂ x u(t, x, y), ∂y u(t, x, y)]
− ∇2 H(u(t, x, y))[∂y u(t, x, y), ∂ x u(t, x, y)],
which of course vanishes. As a result, ω(∂ x u(t, x, y), ∂y u(t, x, y)) is conserved in time.
Furthermore, in the quadratic growth case, in which ∇2 H is bounded, then from the
discussions after Example (1.28), we know that the H-Hamiltonian equation always has
global solutions. Thus, the solution maps S (t) are always well-defined for all t ∈ R.
Elaboration: To show that the solution maps are symplectomorphisms, we must first
provide some additional background detailing how this symplectic form ω is preserved by
the Hamiltonian evolution. Consider the vector space D as a 2n-dimensional real manifold;
recall that each tangent space T x D, where x ∈ D, can be identified with D. 16 Then, we
can impose a symplectic form ω̄ on the manifold D such that at each T x D, the bilinear
form ω̄| x is identified with ω according to the above identification of T x D and D.
For any t ∈ R, the pullback S (t)∗ ω̄ of ω̄ through the solution map S (t) defines a differential form on (the manifold) D. Our goal is to show that S (t)∗ ω̄ = ω̄. Let X, Y be arbitrary
vector fields on D, with coordinate decompositions X = X α ∂α and Y = Y β ∂β with respect
to the standard coordinates. Define also the map
u(t, x) = S (t)x.
u : R × D → D,
Letting dS (t) denote the standard differential map of S (t), then we can compute
S (t)∗ ω̄(X, Y) = ω̄([dS (t)]X, [dS (t)]Y) = X α Y β · ω̄(∂α (yγ ◦ S (t))∂γ , ∂β (yδ ◦ S (t))∂δ ),
where the yγ ’s represent the standard coordinate functions pi and q j . Recalling the pointwise identification between ω̄ and ω, then we have
S (t)∗ ω̄(X, Y)| x = X α Y β ω(∂α (S (t)x), ∂β (S (t)y)) = X α Y β ω(∂α u(t, x), ∂β u(t, x)).
From our conservation property for ω and u, proved in the beginning of this exercise, then
S (t)∗ ω̄(X, Y)| x = X α Y β ω(∂α u(0, x), ∂β u(0, x)) = S (0)∗ ω̄(X, Y) = ω̄(X, Y).
The second to last step follows from the same computations and identifications as before,
and the last step is simply because S (0) is the identity map. As a result, we have shown
that S (t) is indeed a symplectomorphism, as desired.
1.33. Consider the Liouville measure on D defined by the top form
m ≈ ωn = ω ∧ · · · ∧ ω
(n times),
that is, the measure defined
m(Ω) =
Z
Ω
ωn .
Our goal is to show that m(S (t)(Ω)) is the same as m(Ω) for any Lebesgue measurable
Ω ⊆ D. By a change of variables, we have
Z
Z
n
m(S (t)(Ω)) =
ω =
S (t)∗ ωn .
S (t)(Ω)
Ω
By Exercise (1.32) and standard properties of pullback forms, then
S (t)∗ ωn = [S (t)∗ ω]n = ωn ,
16The component vectors q , p ∈ D are identified with the tangent vectors ∂ | , ∂ | ∈ T D, respectively.
i
j
qi x p j x
x
SOLUTIONS
11
and it follows that
m(S (t)(Ω)) =
Z
Ω
ωn = m(Ω).
Finally, since H is conserved by the solution map, then for any t ∈ R,
S (t)∗ [e−βH ωn ] = e−β[H◦S (t)] S (t)∗ ωn = e−β[H◦S (0)] ωn = e−βH ωn .
As a result, letting dµβ = e−βH dm denote the Gibbs measure, we have
Z
Z
Z
dµβ (S (t)(Ω)) =
e−βH ωn =
S (t)∗ [e−βH ωn ] =
e−βH ωn = dµβ (Ω).
Ω
S (t)(Ω)
Ω
1.35. For the Jacobi identity, we break down into “standard” coordinates
(q1 , . . . , qn ; p1 , . . . , pn ),
n = dim D,
described in Exercise 1.28. From computations in Example 1.27, we see that
{H1 , {H2 , H3 }} =
n
X
(∂q j H1 ∂ p j {H2 , H3 } − ∂ p j H1 ∂q j {H2 , H3 })
j=1
=
n X
n
X
(∂q j H1 ∂ p j − ∂ p j H1 ∂q j )(∂ql H2 ∂ pl H3 − ∂ pl H2 ∂ql H3 )
j=1 l=1
=
n
X
(∂q j H1 ∂ p j ql H2 ∂ pl H3 + ∂q j H1 ∂ql H2 ∂ p j pl H3 )
j,l=1
−
n
X
(∂q j H1 ∂ p j pl H2 ∂ql H3 + ∂q j H1 ∂ pl H2 ∂ p j ql H3 )
j,l=1
−
n
X
(∂ p j H1 ∂q j ql H2 ∂ pl H3 + ∂ p j H1 ∂ql H2 ∂q j pl H3 )
j,l=1
+
n
X
(∂ p j H1 ∂q j pl H2 ∂ql H3 + ∂ p j H1 ∂ pl H2 ∂q j ql H3 ).
j,l=1
The brackets {H2 , {H3 , H1 }} and {H3 , {H1 , H2 }} have similar expansions, but with the Hi ’s
permuted. Summing these expansions, we can see that all the individual terms cancel.
Next, for the Leibnitz rule, we first note that
hd(H1 H2 )(u), vi = H1 (u)hdH2 (u), vi + H2 hdH1 (u), vi
= ω(H1 (u)∇ω H2 (u), v) + ω(H2 (u)∇ω H1 (u), v).
In other words, the symplectic gradient satisfies the product rule:
∇ω (H1 H2 ) = H1 ∇ω H2 + H2 ∇ω H1 .
As a result, we can compute as desired
{H1 , H2 H3 } = ω(∇ω H1 , ∇ω H2 )H3 + ω(∇ω H1 , ∇ω H3 )H2
= {H1 , H2 }H3 + H2 {H1 , H3 }.
1.36. We can compute this using the Jacobi identity from Exercise 1.35:
[DH1 , DH2 ]E = {H1 , {H2 , E}} − {H2 , {H1 , E}} = −{E, {H1 , H2 }} = D{H1 ,H2 } E.
12
ARICK SHAO
1.37. First of all, if E and H do not Poisson commute, then by the identity (1.33), there is
a solution curve u to (1.28) for which (E ◦ u)0 = {E, H} ◦ u has constant nonzero sign on
some small interval [t0 , t1 ]. As a result, E(u(t1 )) − E(u(t0 )) is nonzero, but the integral
Z t1
G(u(t))[∂t u(t) − ∇ω H(u(t))]dt
t0
vanishes for any G ∈
→ D∗ ), so that E cannot be an integral of motion.
On the other hand, if E and H do Poisson commute, then on any interval [t0 , t1 ], and for
any C 1 curve u : [t0 , t1 ] → D, we have the identity
Z t1
E(u(t1 )) − E(u(t0 )) =
(E ◦ u)0 (t)dt
0
Cloc
(D
t0
=
Z
=
Z
=
Z
t1
hdE(u(t)), ∂t u(t)idt
t0
t1
[hdE(u(t)), ∂t u(t) − ∇ω H(u(t))i + hdE(u(t)), ∇ω H(u(t))i]dt
t0
t1
hdE(u(t)), ∂t u(t) − ∇ω H(u(t))idt +
t0
Z
t1
{E, H}(u(t))dt.
t0
By our assumption, the last term on the right-hand side vanishes, and it follows that E is
indeed an integral of motion of (1.28).
1.42. We consider the symplectic space (D̄, ω̄), where 17
D̄ = R × R × D,
ω̄((q1 , p1 , u1 ), (q2 , p2 , u2 )) = q1 p2 − p1 q2 + ω(u1 , u2 ).
A quick computation shows that the ω̄-symplectic gradient is given by
∇ω̄ f = (∂ p f, −∂q f, ∇u,ω f ),
f ∈ C 2 (D̄ → R).
In the above, p, q, and u refer to the first, second, and third arguments of f , respectively,
while ∇u,ω f refers to the ω-symplectic gradient of f with respect to the u-variable.
Consider the time-dependent Hamiltonian and the associated Hamiltonian equation
H ∈ C 1 (R × D → R),
∂t u(t) = ∇u,ω H(t, u(t)),
where in the above, “∇u,ω H” refers to the ω-symplectic gradient with respect to the second
argument of H. Consider the following time-independent Hamiltonian on D̄:
H̄ ∈ C 1 (D̄ → R),
H̄(q, p, u) = H(q, u) + p.
A quick computation shows that
∇ω̄ H̄(q, p, u) = (1, −∂q H(q, u), ∇u,ω H(q, u)).
We can now consider the (time-independent) Hamiltonian equation
q(t)
1
∂t p(t) = ∂q H(q(t), u(t)) ,
(q(t0 ), p(t0 ), u(t0 )) = (t0 , 0, u0 ) ∈ D̄.
u(t)
∇u,ω H(q(t), u(t))
where (q, p, u) ∈ C 1 (R → D̄). We can immediately solve the above for q, which yields
q(t) = t. As a result, then u solves the time-dependent Hamiltonian equation
∂t u(t) = ∇u,ω H(t, u(t)).
17In other words, we define ω̄ by combining ω with the symplectic form in Example (1.27).
SOLUTIONS
13
Thus, the above time-dependent Hamiltonian setting can be reformulated as an equivalent
time-independent Hamiltonian setting. Furthermore, note that p satisfies
Z t
Z t
p(t) = −
∂q H(q(s), u(s))ds = −
∂q H(s, u(s))ds.
t0
t0
For such a time-dependent Hamiltonian H, the associated Hamiltonian equation needs
not preserve H. For example, if H(t, u) = t, then we have the equation
∂t u(t) = ∇u,ω t ≡ 0,
u(t0 ) = u0 ,
which has trivial solution u(t) ≡ u0 . However, H fails to be constant in time, since
H(t, u(t)) = H(t, u0 ) = t.
However, by the time-independent Hamiltonian theory, then H̄ is preserved by solution
curves of the H̄-Hamiltonian equation. Thus, a substitute quantity for the time-dependent
H-Hamiltonian equation that is preserved by its solution curves is
Z t
H̄(q(t), p(t), u(t)) = H(t, u(t)) + p(t) = H(t, u(t)) −
∂q H(s, u(s))ds.
t0
1.44. With H defined in terms of L as above, we can first compute the partial derivatives
of H. First of all, q̇ is by definition a function of both p and q, so that
n
n
X
X
∂qi H(q, p) =
(∂qi q̇ j · p j ) − ∂qi L(q, q̇) −
∂q̇ j L(q, q̇) · ∂qi q̇ j
j=1
=
n
X
j=1
∂qi q̇ j · [p j − ∂q̇ j L(q, q̇)] − ∂qi L(q, q̇)
j=1
= −∂qi L(q, q̇).
In the last step, we applied (1.37). By a similar computation, we also have
n
n
X
X
∂ pi q̇ j · ∂q̇ j L(q, q̇) = q̇i .
∂ pi q̇ j · p j −
∂ pi H(q, p) = q̇i +
j=1
j=1
Thus, the Hamiltonian equation is
∂t qi (t) = ∂ pi H(q(t), p(t)) = q̇i (t),
∞
∂t pi (t) = −∂qi H(q(t), p(t)) = ∂qi L(q(t), q̇(t)).
n
Now, fix a curve q ∈ C (I → R ) as in the problem statement. In addition, we define
the “momentum” curve p ∈ C ∞ (I → Rn ) as in (1.37):
p j (t) = ∂q̇ j L(q(t), ∂t q(t)).
Note that in this setup, the first Hamiltonian is trivially satisfied.
Consider now a variation q + εv, with ε sufficiently small, and with v ∈ C ∞ (I → Rn )
vanishing at the endpoints of I. Then, we can compute
Z
d
d
S (q(t) + εv(t))|ε=0 =
L(q(t) + εv(t), ∂t q(t) + ε∂t v(t))dt|ε=0
dε
dε I
n Z
X
=
[∂qi L(q(t), ∂t q(t)) · v(t) + ∂q̇i L(q(t), ∂t q(t)) · ∂t v(t)]dt
i=1
=
I
n Z
X
i=1
[∂qi L(q(t), ∂t q(t)) · v(t) + pi (t) · ∂t v(t)]dt
I
14
ARICK SHAO
Z X
n
=
[−∂t pi (t) + ∂qi L(q(t), ∂t q(t))] · v(t)dt,
I i=1
where in the last step, we integrated by parts to treat the derivative ∂t v(t). Since q is a
critical point of the Lagrangian if and only if the left-hand side vanishes for all such curves
v, then by the above computation, this happens if and only if
∂t pi (t) = ∂qi L(q(t), ∂t q(t)),
1 ≤ i ≤ n.
This is precisely the remaining half of the Hamiltonian equation.
1.45. Consider a variation (q + εv, p + εw) of (q, p), where ε > 0 is small, and where
v, w ∈ C ∞ (I → Rn ) vanish on the endpoints of I. Defining S to be the variation
S=
d
S (q(t) + εv(t), p(t) + εw(t))|ε=0 ,
dε
then we can compute
Z
d
S=
[∂t q(t) + ε∂t v(t)][p(t) + εw(t)] − H(q(t) + εv(t), p(t) + εw(t))dt|ε=0
dε I
Z
= [∂t q(t)w(t) + p(t)∂t v(t) − v(t) · ∇q H(q(t), p(t)) − w(t) · ∇ p H(q(t), p(t))]dt
ZI
= (∂t q(t) − ∇ p H(q(t), p(t)), −∂t p(t) − ∇q H(q(t), p(t))) · (w(t), v(t))dt,
I
where in the last step, we handled the derivative ∂t v(t) by integrating by parts. Therefore,
(q, p) is a critical point of this Lagrangian if and only if the above vanishes for all such v
and w. This happens if and only if the Hamiltonian equations for (q, p):
∂t qi (t) = ∂ pi H(q(t), p(t)),
∂t pi (t) = −∂qi H(q(t), p(t)),
1 ≤ i ≤ n.
This is essentially the inverse to Exercise 1.44. Both exercises assert that one has a critical point with respect to a Lagrangian if and only if the associated Hamiltonian equations
hold. The difference is that Exercise 1.44 states this with respect to L, q, and q̇ = ∂t q,
while Exercise 1.45 states this in terms of H, q, and p. In particular, this demonstrates the
invertible relationships between p and q̇ and between H and L:
p j = ∂q̇ j (q, q̇),
q̇ j = ∂ p j H(q, p),
H(q, p) = q̇ · p − L(q, q̇),
L(q, q̇) = q̇ · p − H(q, p).
1.46. Suppose x is a maximal solution for (1.39), defined on the interval I = (T − , T + ),
which contains t0 = 0. Since V ≥ 0, then for any t ∈ I, we have
p
p
|∂t x(t)| ≤ 2E(t) = 2E(0) < ∞.
Here, we recalled that the energy E(t), defined in (1.40), is conserved.
p
|x(t)| ≤ |x(0)| + |t| sup |∂t x(s)| ≤ |x(0)| + |t| 2E(0) < ∞,
18
Moreover,
t ∈ I.
s∈I
Thus, by Theorem 1.17, we have T ± = ±∞, i.e., x is a C 2 global solution.
Next, suppose x(t0 ) = 0 for some t0 ∈ R, and suppose x(t1 ) = 0 for some t1 , t0 . Since
|x|2 is convex (see Example 1.31), then we obtain for any 0 ≤ α ≤ 1 that
|x(αt0 + (1 − α)t1 )|2 ≤ α|x(t0 )|2 + (1 − α)|x(t1 )|2 = 0.
18See the proof of Proposition 1.24.
SOLUTIONS
15
Thus, x vanishes for all times between t0 and t1 , and by uniqueness, x ≡ 0 everywhere. As
a result, if x does not vanish everywhere, then x can hit zero at at most one point t0 .
Finally, suppose x(t0 ) = 0 for some t0 ∈ R, and let
"
#
|π x(t) ∂t x(t)|2 x(t) · ∇V(x(t))
x(t)
P(t) =
−
= ∂t
· ∂t x(t) = ∂t Q(t),
|x(t)|
|x(t)|
|x(t)|
for any t ∈ R \ {t0 }. 19 By the fundamental theorem of calculus, for large R > 0,
Z R
Z t0 −ε
P(t)dt +
P(t)dt + Q(t0 + ε) − Q(t0 − ε) = Q(R) − Q(−R).
t0 +ε
−R
Since V is radially decreasing, then P ≥ 0, and hence
Z R
Z t0 −ε
√
P(t)dt +
P(t)dt + Q(t0 + ε) − Q(t0 − ε) ≤ 2 sup |Q(s)| ≤ 2 2E.
t0 +ε
s∈R
−R
As the above holds uniformly for all R, then letting R % ∞ yields
Z
√
P(t)dt + Q(t0 + ε) − Q(t0 − ε) ≤ 2 2E.
R\(t0 −ε,t0 +ε)
It remains to compute the limits of Q(t) as t → t0 . First, we have
x(t) · ∂t x(t)
t − t0 x(t)
·
= lim · ∂t x(t) = |∂t x(t)|.
lim
t&0 x(t) t − t0
t&t0
|x(t)|
An analogous calculation yields
x(t) · ∂t x(t)
t − t0 x(t)
·
lim
= − lim · ∂t x(t) = −|∂t x(t)|.
t%t0
t%0 x(t) t − t0
|x(t)|
As a result, letting ε & 0 in the previous inequality, we obtain as desired
Z
√
P(t)dt + 2|∂t x(t0 )| ≤ 2 2E.
R
1.49. Define the map
ϕ : Bε → S,
ϕ(v) = ulin + DN(v).
Note that v ∈ Bε solves (1.50) if and only if v is a fixed point of ϕ. If v ∈ Bε , then
ε
ε 1
kϕ(v)kS ≤ kulin kS + kDN(v)kS ≤ + C0 kN(v) − N(0)kN ≤ + kv − 0kS ≤ ε.
2
2 2
hence ϕ maps Bε into itself. Next, ϕ is a contraction mapping, since for any u, v ∈ Bε ,
1
kϕ(u) − ϕ(v)kS = kDN(u) − DN(v)kN ≤ C0 kN(u) − N(v)kN ≤ ku − vkS .
2
Since Bε is a closed subset of S, it is a complete metric space. By the contraction mapping
theorem, ϕ has a unique fixed point u ∈ B , which is the unique solution in Bε to (1.50).
Lastly, let u, v ∈ Bε denote two solutions to (1.50), with their respective “linear parts”
denoted by ulin , vlin ∈ Bε/2 . Then, by our assumptions, we have the estimate
1
ku − vkS ≤ kulin − vlin kS + kDN(u) − DN(v)kS ≤ kulin − vlin kS + ku − vkS .
2
Rearranging the terms, we obtain
1
ku − vkS ≤ kulin − vlin kS ,
ku − vkS ≤ 2kulin − vlin kS .
2
19See Example 1.32 for details behind this computation.
16
ARICK SHAO
The desired Lipschitz estimate for the “solution map” ulin 7→ u follows. Applying the
above estimate to the special case v = vlin = 0 (the trivial solution) yields
kukS ≤ 2kulin kS .
1.50. Since u = ulin + DN(u), then
ũ − u = DN(ũ) − DN(u) + e.
As a result, by (1.51) and (1.52),
1
kũ − ukS ≤ kekS + kDN(ũ) − DN(u)kS ≤ kekS + kũ − ukS .
2
The desired estimate follows.
1.51. Correction: The correct assumption needed for ε is
1
εk−1 =
.
2kC0C1
We begin by using the triangle inequality and expanding
kN(u) − N(v)kN ≤ kNk (u − v, u, . . . , u)kN + kNk (v, u − v, u, . . . , u)kN
+ · · · + kNk (v, . . . , v, u − v)kN
k−2
k−1
≤ C1 ku − vkS (kukk−1
S + kvkS kukS + · · · + kvkS )
≤ kC1 εk−1 ku − vkS .
Thus, if the above assumption for ε holds, then
kN(u) − N(v)kN ≤
1
ku − vkS .
2C0
1.52. We reduce the second-order ODEs, both homogeneous and nonhomogeneous, to an
equivalent first-order systems by setting v = ∂t u. In other words, we consider the system
∂t u = v, ∂t v = Lu + f ,
u(t0 ) = u0 , v(t0 ) = u1 .
If we define the linear operator
L̃ : D × D → D × D,
L̃(u, v) = (v, Lu)
and the map
f˜ : R → D × D,
then we can rewrite the above system as
∂t (u, v) = L̃(u, v) + f˜,
f˜ = (0, f ),
(u, v)(t0 ) = (u0 , u1 ).
First, consider the linear case f ≡ 0 (i.e., f˜ ≡ 0), from which we have
(u, v)(t) = e(t−t0 )L̃ (u0 , u1 ).
Taking the first component of the above, then we see that there exist operators
Ui : R × D → D,
i ∈ {0, 1},
such that
u(t) = U0 (t − t0 )u0 + U1 (t − t0 )u1 .
More specifically, if we expand exp(t L̃) as a 2 × 2 matrix
"
#
a11 (t) a12 (t)
,
a21 (t) a22 (t)
then U0 (t) = a11 (t) and U1 (t) = a12 (t).
SOLUTIONS
17
Next, for general f , then Duhamel’s principle (Proposition 1.35) yields
Z t
(t−t0 )L̃
(u, v)(t) = e
(u0 , u1 ) +
e(t−s)L̃ (0, f )ds.
t0
Projecting to the first component, and with U0 and U1 as before, we obtain
Z t
u(t) = U0 (t − t0 )u0 + U1 (t − t0 )u1 +
U1 (t − s) f (s)ds.
t0
1.54. First, we take a derivative of ku(t)k2 in order to obtain the inequality
∂t ku(t)k2 = 2h∂t u, ui = 2hLu, ui ≤ −2σku(t)k2 .
Applying the differential form of the Gronwall inequality, we have
!
Z t
2
2
ku(t)k ≤ ku(0)k exp −2
σds = e−2σt ku(0)k2 .
0
The desired inequality follows immediately.
1.57. Correction: In contrast to the problem statement, the ODE for φ we wish to solve is
∂t φ(t) = −P(u(t))φ(t),
φ(t0 ) = φ0 .
In particular, note the change in sign in the right-hand side of the ODE.
First, a solution φ exists, as it can be given explicitly using matrix exponentials:
" Z t
#
φ(t) = exp −
P(u(s))ds · φ0 .
t0
This solution is also unique, since if φ and ψ are both solutions, with the same initial
condition φ0 at time t0 , then their difference α = ψ − φ satisfies
∂t α(t) = −P(u(t))α(t),
α(t0 ) = 0.
From this, we immediately obtain the estimate
∂t |α(t)|2 = −2hP(u(t))α(t), α(t)i ≤ 2|P(u(t))||α(t)|2 ,
and Gronwall’s inequality ensures that α vanishes for all time.
Consider now the curve
t 7→ v(t) = L(u(t))φ(t) − λφ(t)
in H, and note first that v(t0 ) = 0. Differentiating this curve and recalling both the above
ODE and the definition of Lax pairs, we have
∂t v(t) = ∂t [L(u(t))] · φ(t) + [L(u(t)) − λ]∂t φ(t)
= (LP − PL − LP + λP)|u(t) φ(t) = −P(u(t))v(t).
In other words, v satisfies the ODE in the previous paragraph, with vanishing initial data.
Therefore, v vanishes everywhere as well by the previous argument. In particular, this
shows that the spectrum σ(L) of L is an invariant of the flow (1.57) of u.
18
ARICK SHAO
Chapter 2: Constant Coefficient Linear Dispersive Equations
2.1. Let z = (z1 , z2 , z3 , z4 ) ∈ V. The commutation relations are brute force calculations:
1
1
γ0 γ0 z = 2 (z1 , z2 , z3 , z4 ),
γ0 γ1 z = (z4 , z3 , z2 , z1 ),
c
c
1
i
γ0 γ3 z = (z3 , −z4 , z1 , −z2 ),
γ0 γ2 z = (−z4 , z3 , −z2 , z1 ),
c
c
1
1 1
1 0
γ γ z = (−z4 , −z3 , −z2 , −z1 ),
γ γ z = (−z1 , −z2 , −z3 , −z4 ),
c
γ1 γ2 z = i(−z1 , z2 , −z3 , z4 ),
γ1 γ3 z = (z2 , −z1 , z4 , −z3 ),
i
γ2 γ1 z = i(z1 , −z2 , z3 , −z4 ),
γ2 γ0 z = − (z4 , −z3 , z2 , −z1 ),
c
γ2 γ2 z = (−z1 , −z2 , −z3 , −z4 ),
γ2 γ3 z = i(−z2 , −z1 , −z4 , −z3 ),
1
γ3 γ0 z = (−z3 , z4 , −z1 , z2 ),
γ3 γ1 z = (−z2 , z1 , −z4 , z3 ),
c
γ3 γ2 z = i(z2 , z1 , z4 , z3 ),
γ3 γ3 z = (−z1 , −z2 , −z3 , −z4 ).
As a result, we can check every possibility for (γα γβ + γβ γα )z:
2
(γ0 γ0 + γ0 γ0 )z = 2 z,
(γ0 γ1 + γ1 γ0 )z = 0,
c
(γ0 γ2 + γ2 γ0 )z = 0,
(γ0 γ3 + γ3 γ0 )z = 0,
(γ1 γ1 + γ1 γ1 )z = −2z,
(γ1 γ2 + γ2 γ1 )z = 0,
(γ1 γ3 + γ3 γ1 )z = 0,
(γ2 γ2 + γ2 γ2 )z = −2z,
(γ2 γ3 + γ3 γ2 )z = 0,
(γ3 γ3 + γ3 γ3 )z = −2z.
To see the symmetry of the γα ’s, we let w = (w1 , w2 , w3 , w4 ) ∈ V as well. Then,
1
1
{γ0 z, w} = {(z1 , z2 , −z3 , −z4 ), (w1 , w2 , w3 , w4 )} = (z1 w̄1 + z2 w̄2 + z3 w̄3 + z4 w̄4 ),
c
c
1
1
{z, γ0 w} = {(z1 , z2 , z3 , z4 ), (w1 , w2 , −w3 , −w4 )} = (z1 w̄1 + z2 w̄2 + z3 w̄3 + z4 w̄4 ),
c
c
1
{γ z, w} = {(z4 , z3 , −z2 , −z1 ), (w1 , w2 , w3 , w4 )} = (z4 w̄1 + z3 w̄2 + z2 w̄3 + z1 w̄4 ),
{z, γ1 w} = {(z1 , z2 , z3 , z4 ), (w4 , w3 , −w2 , −w1 )} = (z1 w̄4 + z2 w̄3 + z3 w̄2 + z4 w̄1 ),
{γ2 z, w} = i{(−z4 , z3 , z2 , −z1 ), (w1 , w2 , w3 , w4 )} = i(−z4 w̄1 + z3 w̄2 − z2 w̄3 + z1 w̄4 ),
{z, γ2 w} = −i{(z1 , z2 , z3 , z4 ), (−w4 , w3 , w2 , −w1 )} = −i(−z1 w̄4 + z2 w̄3 − z3 w̄2 + z4 w̄1 ),
{γ3 z, w} = {(z3 , −z4 , −z1 , z2 ), (w1 , w2 , w3 , w4 )} = (z3 w̄1 − z4 w̄2 + z1 w̄3 − z2 w̄4 ),
{z, γ3 w} = {(z1 , z2 , z3 , z4 ), (w3 , −w4 , −w1 , w2 )} = (z1 w̄3 − z2 w̄4 + z3 w̄1 − z4 w̄2 ).
Direct inspection of the above formulas establishes symmetry. In particular, note that
{z, γ0 z} = c−1 |z|2 ≥ 0.
To show the final positivity property, we painfully expand expressions. First,
{z, z}2 = (|z1 |2 + |z2 |2 − |z3 |2 − |z4 |2 )2
= |z1 |4 + |z2 |4 + |z3 |4 + |z4 |4 + 2|z1 |2 |z2 |2 − 2|z1 |2 |z3 |2
− 2|z1 |2 |z4 |2 − 2|z2 |2 |z3 |2 − 2|z2 |2 |z4 |2 + 2|z3 |2 |z4 |2 .
SOLUTIONS
19
Next, since γ0 z = c(−z1 , −z2 , z3 , z4 ), then
−{z, γ0 z}{z, γ0 z} = (|z1 |2 + |z2 |2 + |z3 |2 + |z4 |2 )2 ,
= |z1 |4 + |z2 |4 + |z3 |4 + |z4 |4 + 2|z1 |2 |z2 |2 + 2|z1 |2 |z3 |2
+ 2|z1 |2 |z4 |2 + 2|z2 |2 |z3 |2 + 2|z2 |2 |z4 |2 + 2|z3 |2 |z4 |2 .
Moreover, since the remaining γi ’s are identical to the γi ’s, then 20
−{z, γ1 z}{z, γ1 z} = −(z4 z̄1 + z3 z̄2 + z2 z̄3 + z1 z̄4 )2 = −4[R(z1 z̄4 + z3 z̄2 )]2 = −A2 ,
−{z, γ2 z}{z, γ2 z} = −(−z4 z̄1 + z3 z̄2 − z2 z̄3 + z1 z̄4 )2 = −4[I(z1 z̄4 + z3 z̄2 )]2 = −B2 ,
−{z, γ3 z}{z, γ3 z} = −(z1 z̄3 − z2 z̄4 + z3 z̄1 − z4 z̄2 )2 = −4[R(z1 z̄3 − z2 z̄4 )]2 = −C 2 ,
Combining all the above, then we must show
4|z1 |2 |z4 |2 + 4|z2 |2 |z3 |2 + 4|z1 |2 |z3 |2 + 4|z2 |2 |z4 |2 − A2 − B2 − C 2 ≥ 0.
By a direct calculation while tracking cancellations, then
−A2 − B2 = −4|z1 z̄4 + z3 z̄2 |2 = −4|z1 |2 |z4 |2 − 4|z3 |2 |z2 |2 − 8R(z1 z2 z̄3 z̄4 ),
−C 2 = −4[R(z1 z̄3 )]2 − 4[R(z2 z̄4 )]2 + 8R(z1 z̄3 )R(z2 z̄4 ).
Since
R(ab) = Ra · Rb − Ia · Ib
for all a, b ∈ C, then applying Cauchy’s inequality yields
−A2 − B2 − C 2 = −4|z1 |2 |z4 |2 − 4|z3 |2 |z2 |2 − 4[R(z1 z̄3 )]2 − 4[R(z2 z̄4 )]2 + 8I(z1 z̄3 )I(z2 z̄4 )
≥ −4|z1 |2 |z4 |2 − 4|z3 |2 |z2 |2 − 4[R(z1 z̄3 )]2 − 4[R(z2 z̄4 )]2
− 4[I(z1 z̄3 )]2 − 4[I(z2 z̄4 )]2
= −4|z1 |2 |z4 |2 − 4|z3 |2 |z2 |2 − 4|z1 |2 |z3 |2 − 4|z2 |2 |z4 |2 .
This completes the proof of the timelike property.
2.2. First, taking a time derivative of the Maxwell equations yields
∂2t E = c2 ∇ x × ∂t B = −c2 ∇ x × ∇ x × E = c2 [∆ x E − ∇ x (∇ x · E)] = c2 ∆ x E,
∂2t B = −∇ x × ∂t E = −c2 ∇ x × ∇ x × B = c2 [∆ x B − ∇ x (∇ x · B)] = c2 ∆B,
Thus, all components of E and B satisfy the wave equation. Next, for the abelian YangMills equations, we take a spacetime divergence of the second equation in (2.6) to obtain
0 = ∂α ∂α Fβγ + ∂α ∂β Fγα + ∂α ∂γ Fαβ = Fβγ − ∂β (∂α Fαγ ) + ∂γ (∂α Fαβ ) = Fβγ .
In particular, the spacetime divergence of F vanishes due to the first equation in (2.6).
Correction: In order to show that a solution
2
(R1+d → R1+d )
A ∈ Ct,x
of the wave equation can also be reformulated as a solution of the abelian Yang-Mills
equations, we must assume in addition the Lorenz gauge condition
∂α Aα ≡ 0.
With A as above, we define the “curvature” two-form
Fαβ = ∂α Aβ − ∂β Aα .
20Here, R and I refer to the real and imaginary components, respectively.
20
ARICK SHAO
A direct computation now yields
∂α Fαβ = Aβ − ∂β ∂α Aα ≡ 0.
Furthermore, the definition of F yields the Bianchi identities:
∂α Fβγ + ∂β Fγα + ∂γ Fαβ = ∂α ∂β Aγ − ∂α ∂γ Aβ + ∂β ∂γ Aα − ∂β ∂α Aγ + ∂γ ∂α Aβ − ∂γ ∂β Aα ≡ 0.
Thus, F satisfies the abelian Yang-Mills equations.
We now restrict ourselves to the case d = 3. To see that the Maxwell equations are
a special case of the abelian Yang-Mills equations, we let F be a solution of the abelian
Yang-Mills equations, and we define E and H in terms of F by
E1 = F10 ,
E2 = F20 ,
E3 = F30 ,
H1 = F23 ,
H2 = F31 ,
H3 = F12 .
From the first equation in (2.6), we can compute that
0 ≡ ∂1 F10 + ∂2 F20 + ∂3 F30 = ∇ x · E,
0 ≡ ∂0 F01 + ∂2 F21 + ∂3 F31 = c−2 ∂t E1 − (∇ x × H)1 ,
0 ≡ ∂0 F02 + ∂1 F12 + ∂3 F32 = c−2 ∂t E2 − (∇ x × H)2 ,
0 ≡ ∂0 F03 + ∂1 F13 + ∂2 F23 = c−2 ∂t E3 − (∇ x × H)3 .
Similarly, for the second equation in (2.6), we can compute
0 ≡ ∂0 F12 + ∂1 F20 + ∂2 F01 = ∂t H3 + (∇ x × E)3 ,
0 ≡ ∂0 F13 + ∂1 F30 + ∂3 F01 = −∂t H2 − (∇ x × H)2 ,
0 ≡ ∂0 F23 + ∂2 F30 + ∂3 F02 = ∂t H1 + (∇ x × E)1 ,
0 ≡ ∂1 F23 + ∂2 F31 + ∂3 F12 = ∇ x · H.
As a result, we have recovered the Maxwell equations.
Next, if u satisfies the Dirac equations, then we have
mc
m2 c2
u = i γβ ∂β u = −γβ ∂β (γα ∂α u) = −γα γβ (∂α ∂β u).
~
~2
Since partial derivatives commute, then
m2 c2
1
u = − (γα γβ + γβ γα )∂α ∂β u = gαβ ∂α ∂β u = u,
2
~2
so u also satisfies the Klein-Gordon equations.
Correction: For the converse, if φ satisfies the Klein-Gordon equations, then we show
ψ = γ α ∂α φ − i
mc
φ
~
satisfies the Dirac equation. Note the extra factor −ic~−1 required on the right-hand side.
With φ as above, then a direct computation yields
iγβ ∂β ψ = −iφ +
mc β
m2 c 2
mc β
mc
γ ∂β φ =
γ ∂β φ − i 2 φ =
· ψ.
~
~
~
~
Thus, ψ indeed satisfies the Dirac equation.
SOLUTIONS
21
2.5. Correction: Define the Galilean transformed function ũ by 21
E(t, x) = eimx·v/~ e−imt|v|
2
/2~
ũ(t, x) = E(t, x)u(t, x − vt).
,
Direct computations yield
X
−im|v|2
j
∂t ũ(t, x) = E(t, x)
u(t, x − vt) + ∂t u(t, x − vt) −
v ∂ j u(t, x − vt) ,
2~
j
"
#
j
imv
∂ j ũ(t, x) = E(t, x)
u(t, x − vt) + ∂ j u(t, x − vt) ,
~
X 2imv j
−m2 |v|2
u(t, x − vt) + ∆u(t, x − vt) +
∆ũ(t, x) = E(t, x)
∂ j u(t, x − vt) ,
2
~
~
j
Combining the above equations, we obtain
!
!
~
~
i∂t +
∆ ũ(t, x) = E(t, x) i∂t +
∆ u(t, x − vt).
2m
2m
Since |E| ≡ 1 (in particular, E is always nonvanishing), it follows that ũ solves the linear
Schrödinger equations if and only if u does.
2.9. Suppose u : R × Rd → C is a solution of (2.1), and define
uλ (t, x) = u(λ−k t, λ−1 x),
uλ : R × Rd → C,
where k is the degree of L and P. Expanding
X
X
pα ξα ,
L = P(∇) =
pα ∂αx ,
P(ξ) =
|α|=k
pα ∈ C,
|α|=k
then we can compute
∂t uλ (t, x) = ∂t [u(λ−k t, λ−1 x)] = λ−k ∂t u(λ−k t, λ−1 x) = λ−k Lu(λ−k t, λ−1 x).
We can similarly compute Luλ via the chain rule:
X
X
Luλ (t, x) =
pα ∂αx [u(λ−k t, λ−1 x)] =
λ−k pα ∂αx u(λ−k t, λ−1 x) = λ−k Lu(λ−k t, λ−1 x).
|α|=k
|α|=k
The above shows that ∂t uλ and Luλ are the same, so that uλ solves (2.1).
2.17. Applying the spatial Fourier transform of the transport equation
∂t u(t, x) = −x0 · ∇u(t, x),
then we obtain
∂t û(t, ξ) = −i(x0 · ξ)û(t, ξ).
Solving this ODE with respect to t yields
û(t, ξ) = e−(x0 ·ξ)t û0 (ξ) = e−(tx0 ·ξ) û0 (ξ),
and taking an inverse Fourier transform yields
u(t, x) = u0 (x − tx0 ).
As a result,
exp(−tx0 · ∇) f (x) = f (x − tx0 ),
21Note the difference in sign in the second exponent.
exp(−x0 · ∇) f (x) = f (x − x0 ).
22
ARICK SHAO
If f is real analytic, then we can solve for a solution u to the transport equation that is
also real analytic in the time variable. We begin by assuming the ansatz
X
u(t, x) =
ak (x) · tk ,
u(0, x) = a0 (x) = f (x).
k
The transport equation applied termwise to the summation yields
X
X
(k + 1)ak+1 (x) · tk =
∇−x0 ak (x) · tk .
k
k
In other words, for each k ≥ 0, we must solve
1
ak+1 (x) =
∇−x0 ak (x).
k+1
Since a0 = f , then by induction, we can show that 22
1
1
ak (x) = (∇−x0 )k f (x) = (∇− x0 )k f (x) · |x0 |k ,
k ≥ 0.
|x0 |
k!
k!
Plugging this in and recalling Taylor’s formula, since f is real analytic, we obtain
X 1
(∇ x0 )k f (x) · (|x0 |t)k = f (x − x0 t).
u(t, x) =
k! − |x0 |
k
2.18. The spatial Fourier transform of the wave equation is
∂t û(t, ξ) = −|ξ|2 û(t, ξ),
which, as an ODE in t, has general solution
û(t, ξ) = a(ξ) cos(t|ξ|) + b(ξ) sin(t|ξ|).
Setting t = 0 yields û0 (ξ) = a(ξ). Next, differentiating the above in time yields
∂t û(t, ξ) = −a(ξ)|ξ| sin(t|ξ|) + b(ξ)|ξ| cos(t|ξ|).
Setting t = 0 yields û1 (ξ) = |ξ|b(ξ). Thus, the wave equation has a solution
sin(t|ξ|)
· û1 (ξ).
û(t, ξ) = cos(t|ξ|) · û0 (ξ) +
|ξ|
Since the operator√−∆ in physical space corresponds to multiplying by the factor |ξ|2 in
Fourier space, then −∆ corresponds to multiplication by |ξ|. Thus, in terms of Fourier
multipliers, we can write the above solution in physical space as
√
√
sin(t −∆)
u(t) = cos(t −∆) · u0 +
· u1 .
√
−∆
For the spacetime Fourier transform, we take a Fourier transform in time of the above
representation formula for û. Recalling the standard formulas for the Fourier transforms of
the functions t 7→ cos(at) and t 7→ sin(at), we have
π
[δ(τ − |ξ|) − δ(τ + |ξ|)] · û1 (ξ).
ũ(τ, ξ) = π[δ(τ − |ξ|) + δ(τ + |ξ|)] · û0 (ξ) +
i|ξ|
Somewhat informally, since δ(τ − |ξ|), δ(τ + |ξ|), and δ(|τ| − |ξ|) correspond to integrals over
the upper null cone, the lower null cone, and the full null cone beginning at the origin,
respectively, then one can derive the identities
δ(τ − |ξ|) + δ(τ + |ξ|) = δ(|τ| − |ξ|),
δ(τ − |ξ|) − δ(τ + |ξ|) = δ(|τ| − |ξ|) sgn τ.
22In the special case x = 0, then solving the above equations for the a ’s yields that a ≡ 0 for all k > 0, so
0
k
k
that u(t, x) = f (x) for all t ∈ R and x ∈ Rd , as desired.
SOLUTIONS
23
For the second formula, one notes that the hyperplanes {τ − |ξ| = 0} and {τ + |ξ| = 0}
correspond to the portions of {|τ| − |ξ| = 0} with τ > 0 and τ < 0, respectively. As a result,
we obtain the desired formula
"
#
1
sgn(τ)
ũ(τ, ξ) = 2π · δ(|τ| − |ξ|) û0 (ξ) +
û1 (ξ) .
2
2i|ξ|
Correction: The above spacetime Fourier identity for the wave equation differs from
the problem statement by a factor of 2π.
For the H s -estimates, we use the Plancherel theorem and the above Fourier identity:
k∇u(t)kHxs−1 . k(1 + |ξ|2 )
s−1
2
|ξ||û0 |2 kLξ2 + k(1 + |ξ|2 )
s−1
2
|û1 |2 kLξ2 . ku0 kHxs + ku1 kHxs−1 .
Recall that cos(t|ξ|) and sin(t|ξ|) are uniformly bounded by 1. Repeating this process with
the Fourier identity for ∂t û (and replacing a and b as before), then we obtain the bound
k∂t u(t)kHxs−1 . k(1 + |ξ|2 )
s−1
2
|ξ||û0 |2 kLξ2 + k(1 + |ξ|2 )
s−1
2
|û1 |2 kLξ2 . ku0 kHxs + ku1 kHxs−1 .
For the lower order bounds, again using the spatial Fourier representations, we obtain
ku(t)kHxs . k∇u(t)kHxs−1 + ku(t)kHxs−1
. ku0 kHxs + ku1 kHxs−1 + k(1 + |ξ|2 )
s−1
2
û0 kLξ2 + k(1 + |ξ|2 )
. ku0 kHxs + ku1 kHxs−1 + k(1 + |ξ|2 )
s−1
2
|ξ|−1 sin(t|ξ|)û1 kLξ2 .
s−1
2
|ξ|−1 sin(t|ξ|)û1 kLξ2
If we write the sine factor as
Z
| sin(t|ξ|)| .
t
|ξ|| sin(s|ξ|)|ds . t|ξ|,
0
then we obtain the following control:
ku(t)kHxs . ku0 kHxs + ku1 kHxs−1 + tk(1 + |ξ|2 )
s−1
2
û1 kLξ2
. hti(ku0 kHxs + ku1 kHxs−1 ).
This proves the last identity in the problem statement.
2.21. Define the quantity
v(t) = e(t0 −t)L u(t),
and note that
∂t v(t) = −Le(t0 −t)L u(t) + e(t0 −t)L ∂t u(t) = e(t0 −t)L F(t).
Integrating the above and then applying the propagator exp[(t − t0 )L] yields
Z t
Z t
e(t0 −t)L u(t) = u(t0 ) +
e(t0 −s)L F(s)ds,
u(t) = e(t−t0 )L u0 +
e(t−s)L F(s)ds.
t0
t0
2.25. Let I = [a, b]. Taking the spatial Fourier transform of the wave equation for u yields
∂2t û(t, ξ) = −|ξ|2 û(t, ξ).
By the Plancherel theorem,
Z
u(t, x)dt
I
Ḣ x2
Z
' |ξ|2 û(t, ξ)dt .
I
Lξ2
By the above Fourier wave equation for û and the Plancherel theorem, then
Z
Z
u(t, x)dt ' ∂2 û(t, ξ)dt ≤ k∂ û(b, ξ)k 2 + k∂ u(a, ξ)k 2 .
t
t
Lξ
Lξ
t
2 2
I
Ḣ x
I
Lξ
24
ARICK SHAO
Applying the energy estimate for the wave equation (see Exercise (2.17)), we have
Z
u(t, x)dt . k∂ u(0)k 2 ≤ ku(0)k 1 + k∂ u(0)k 2 .
t
t
Lx
Lx
Ḣ x
2
I
2.29.
23
Ḣ x
Let I = [a, b]. For the first estimate, we integrate by parts:
Z
Z b
1
eiφ(x) dx =
∂ [eiφ(x) ]dx
0 (x) x
iφ
I
a
#
Z b "
1
1
1
∂x 0
eiφ(x) dx
= 0 eiφ(b) − 0 eiφ(a) +
iφ (b)
iφ (a)
φ
(x)
a
= I1 + I2 + I3 .
By the assumption |φ0 | ≥ λ, we see that 1/φ0 has constant sign throughout I, so that
"
#
1
1
|I1 + I2 | ≤ ± 0
+ 0
,
φ (b) φ (a)
for the correctly chosen sign. Next, since φ00 has constant sign, then
#
"
#
Z b "
1
1
1
dx = ± 0
− 0
,
|I3 | ≤ ±
∂x 0
φ (x)
φ (b) φ (a)
a
again for the correctly closen sign. Adding the estimate for |I1 + I2 | to that for |I3 |, we see
that one pair of terms have to cancel, so that only one pair remains. As a result,
"
#
Z
1
1
2
eiφ(x) dx ≤ 2 max
,
≤ .
|φ0 (b)| |φ0 (a)|
λ
I
For general k > 1, we use an induction argument. First, the base case k = 1 is proved
by the above. Suppose now that the desired estimate
Z
eiφ(x) dx . λ− 1k ,
φ ∈ C 2 (I),
|∂kx φ| ≥ λ
k
I
holds for the case k, and consider the case k + 1. 24
k+1
k
Let φ ∈ C k+1 (I) satisfy |∂k+1
x φ| ≥ λ, and fix δ > 0. By this assumption on ∂ x φ = ∂ x ∂ x φ,
k
then the uniform lower bound |∂ x φ| ≥ δλ must hold everywhere on I except possibly for a
subinterval I0 of length at most 2δ. Partition I \ I0 into subintervals
I + = {x ∈ I | x > y for all y ∈ I0 },
I − = {x ∈ I | x < y for all y ∈ I0 }.
Applying the induction hypothesis, we have 25
Z
Z
+ . (δλ)− 1k + (δλ)− 1k . (δλ)− 1k .
iφ(x)
iφ(x)
e
dx
e
dx
−
+
k
I
I
Next, for I0 , we have the trivial estimate
Z
iφ(x)
e dx ≤ 2δ.
I0
As a result, we have
Z
eiφ(x) dx . (δλ)− 1k + δ.
k
I
23The solution was obtained partially from [4].
24In the case k = 1, we must also assume that φ is convex or concave.
25Note that if k = 1, then our assumption |∂1+1 φ| ≥ λ automatically implies that φ is convex or concave.
x
SOLUTIONS
25
Optimizing the inequality by choosing δ ∼ λ−1/(k+1) , then we obtain as desired
Z
1
eiφ(x) dx . λ− k+1
.
k+1
I
Finally, if ψ is a function on I of bounded variation (so that it is differentiable a.e.), then
Z x
Z
Z
eiφ(x) ψ(x)dx = ∂
iφ(y)
e
dy
·
ψ(x)dx
x
I
a
I
Z b
Z Z x
iφ(y)
iφ(y)
0
e dy · ψ(b) −
e dy · ψ (x)dx
=
a
I a
"
#
Z
1
.k λ− k |ψ(b)| + |ψ0 (x)|dx .
I
where in the last step, we applied the previous estimates for the integral of eiφ(x) .
2.35. Suppose the given estimate holds for all u ∈ S x (Rd ), with Ct being the optimal
constant (i.e., the operator norm) for a given t. Given u0 ∈ S x and λ > 0, we define
uλ0 ∈ S x ,
uλ0 (x) = u0 (λ−1 x).
By a change of variables, then we obtain the first inequality
keit∆/2 uλ0 kLqx ≤ Ct tα kuλ0 kLxp = Ct tα λ p ku0 kLxp .
d
The rescaling property of Exercise (2.9) implies the identity
eit∆/2 uλ0 (x) = ei·λ
−2
t·∆/2
u0 (x/λ),
so that by a similar change of variables, we have the second inequality
keit∆/2 uλ0 kLqx = λ q kei·λ
d
−2
t·∆/2
u0 kLqx ≤ Cλ−2 t tα λ q −2α ku0 kLxp .
d
By choosing u0 that almost fulfills the constant Ct in the first inequality above, then dividing
the second inequality by the first yields
d
d
λ q − q −2α & 1.
By reversing the roles of the above inequalities, we also have
λ p +2α− q & 1.
d
d
By varying λ over all positive real numbers, it is clear then that
!
d d
d 1 1
− − 2α = 0,
α=
−
.
q p
2 q p
Next, combining the above and Exercise (2.34), we have for any u0 ∈ S x (Rd ) that
d 1
1
keit∆/2 u0 kLqx ≤ Ct t 2 ( q − p ) ku0 kLxp ,
1
1
keit∆/2 u0 kLqx 'u0 htid( q − 2 ) .
Choose u0 such that the constant C in the first inequality is almost realized. Dividing each
inequality by the other, as before, and varying t over large positive real numbers, we obtain
!
!
1 1
d 1 1
d
−
=
−
,
q 2
2 q p
and by simple algebra this yields q = p0 .
Finally, if q < p, then
!
d 1 1
−
> 0,
2 q p
26
ARICK SHAO
so that for any u0 ,
d 1
1
lim keit∆/2 u0 kLqx .u0 lim t 2 ( q − p ) = 0.
t&0
t&0
However, this contradicts Exercise (2.34), which implies that
1
1
1
1
lim htid( q − 2 ) = 1.
keit∆/2 u0 kLqx 'u0 htid( q − 2 ) ,
t&0
As a result, we have that q ≥ p.
2.42. Suppose that the Strichartz inequality (2.24) holds for some p, q, d. If u solves the
linear Schrödinger equation, and if λ > 0, then
uλ (t, x) = u(λ−2 t, λ−1 x)
is also a solution of the linear Schrödinger equation, so that
λ > 0.
kuλ kLtq Lrx . kuλ (0)kL2x ,
By a simple change of variables, we see that
kuλ kLtq Lrx = λ r + q kukLtq Lrx ,
2
d
This shows that
d
kuλ (0)kL2x = λ 2 ku(0)kL2x .
λ r + q − 2 kukLtq Lrx . ku(0)kL2x ,
d
2
d
λ > 0,
independently of λ. This can only hold if
d 2 d
+ = .
r q 2
Next, let u be a Schwartz solution of the linear Schrödinger equation, and fix a sequence
t1 < t2 < t3 < . . . ,
with the tn ’s spaced “sufficiently far” apart. Moreover, for any integer N > 0, we define
uN (t) =
N
X
u(t − tn ),
n=1
which also solves the linear Schrödinger equation. We can estimate uN (0) in L2 as follows:
kuN (0)k2L2 =
N
X
x
ke−iti ∆ u(0)k2L2 +
X
x
he−iti ∆ u(0), e−it j ∆ u(0)i
i, j
i=1
= Nku(0)k2L2 +
XZ
x
i, j
2
ei(t j −ti )|ξ| |û(0, ξ)|2 dξ.
Rd
Via stationary phase methods, if the tn ’s are spaced sufficiently far apart, then
N X
∞ Z
X
2
2
kuN (0)kL2 .u(0) N +
ei(t j −ti )|ξ| |û(0, ξ)|2 dξ .u(0) N.
x
i=1 j=1
Rd
1
As a result, we have obtained kuN (0)kL2x .u(0) N 2 .
Next, we consider the Ltq Lrx -norm of uN . Let Bn denote the interval (tn −, tn +) for some
small enough so that each Bn is very far away from all the other ti ’s. By our assumptions
on q and r, then by Exercise (2.34),
2
ku(t − tn )kLr 'u,d,r ht − tn i− q .
SOLUTIONS
27
In particular, since the tn ’s are spaced very far apart, then for any on t ∈ Bn , we have
X
ku(t − tn )kLrx ku(t − ti )kLrx ,
kuN (t)kLr & ku(t − tn )kLrx .
1≤i≤∞
i,n
Therefore, we can estimate from below as follows:
kuN kLtq Lrx
N Z
q1
X
≥
kuN (t)kqLrx dt
Bn
n=1
N Z
q1
X
q
&
ku(t − tn )kLrx dt
Bn
n=1
= N
Z
! 1q
ku(t)kLrx dt
−
1
& Nq.
In particular, if the Strichartz estimates hold for the above values of q, r, and d, then
q ≥ 2, since otherwise, taking N % ∞, we see that the lower bound for kuN (0)kL2x grows
faster than the upper bound for kuN kLtq Lrx .
2.46. First of all, the Littlewood-Paley and integral Minkowski inequalities imply that
12
21 X
X
2
2
ku(t)kLrx 'r,d |PN u(t)| ≤ kPN u(t)kLrx
N
N
Lr
x
for any t. Therefore, applying Minkowski’s inequality again, we obtain
kukLtq Lrx (I×Rd ) .r,d
Z
q2 q1
X
kP u(t)k2 r dt
N
Lx
I
N
! 2q 21
X Z
q
≤
kPN u(t)kLrx dt
N
I
X
= kPN uk2 q
N
Lt
21
.
Lr (I×Rd )
x
For the “dual” estimate, we again apply the integral Minkowski inequality to obtain
X
kPN uk2 q0
Lt
N
1
21 21 Z
q20 q0
X
X
.
≤ kPN uk2Lr0 dt ≤ |PN u|2 0
Lrx (I×Rd )
0 0
x
I
q r
N
N
d
L L (I×R )
t
x
Another application of the Littlewood-Paley inequality yields the desired dual inequality.
Finally, for the Besov Strichartz inequality, we apply Theorem (2.3) to obtain
21
12
12
X it∆/2
X
2
2
PN u0 kLq Lr . kPN u0 kL2 . ku0 kL2x ,
= ke
Lt Lrx
x
t x
X
kPN eit∆/2 u0 k2 q
N
N
N
28
ARICK SHAO
since each PN commutes with i∂t + ∆, and hence its linear propagator. Similarly, we have
21
Z
2 21
Z
X
X
e−is∆/2 F(s)ds . e−is∆/2 P F(s)ds . 0 0 kP F(s)k2
,
N
N
q̃0 r̃0
d,q̃ ,r̃
L2x
R
L2x
R
N
N
Lt L x
where in the last step, we applied (2.25). Finally, for the Besov inhomogeneous Strichartz
estimate, we apply (2.26) in order to obtain for any band N that
Z
i(t−s)∆/2
e
PN F(s)ds
.d,q,r,q̃0 ,r̃0 kPN F(s)kLq̃0 Lr̃0 .
x
t
q r
s<t
Lt L x
Taking an ` -summation of the above over N yields the Besov analogue of (2.26).
2
2.47. First, we compute the symplectic gradient ∇ω H. Since for any “nice” u, v ∈ L2 (Rd ),
Z
Z
Z
d
1 d
2 Re(∇u · ∇v) = −
|∇u + ε∇v| =
H(u + εv)|ε=0 =
Re(∆u · v̄),
d
dε
2 dε Rd
Rd
ε=0
Z
Z R
ω(∇ω H(u), v) = −2
Im(∇ω H(u) · v̄) = 2
Re(i∇ω H(u) · v̄),
Rd
Rd
then varying v appropriately, we obtain
2∇ω H(u) = i∆u.
Thus, Hamilton flow associated to H is the linear Schrödinger equation,
1
i∂t u = i∇ω H(u) = − ∆u.
2
Since {H, H} ≡ 0 trivially, then H is conserved on the Hamilton flow of H, i.e.,
1
i∂t u = − ∆u.
2
This is the conservation of energy. The corresponding symmetry from Noether’s theorem is
given by by flow along the Hamiltonian equation for H. Since integrating along this flow
produces time translates of solutions to the H-Hamiltonian (i.e., Schrödinger) equation,
then the corresponding symmetries for H are time translations.
Next, consider the total mass/probability
Z
M(u) =
|u|2 ,
u ∈ L2 (Rd ).
H(u(t)) = H(u(0)),
Rd
2
d
Since for any “nice” u, v ∈ L (R ), we have
Z
Z
d
M(u + εv)|ε=0 = 2
Re(u · v̄),
ω(∇ω M(u), v) = 2
Re(i∇ω M(u) · v̄),
dε
Rd
Rd
then we have ∇ω M(u) = −iu. The associated M-Hamiltonian equation is
∂t u(t) = −iu(t),
which has solution flows
u(t) = e−it u(0).
Note that H is clearly conserved along this flow. By Noether’s theorem, we have
1
i∂t u = − ∆u,
2
which is the conservation of mass/probability. Furthermore, the corresponding symmetry
for H is given by the solutions of the M-flows, i.e., the phase rotations.
M(u(t)) = M(u(0)),
SOLUTIONS
29
For the momentum functionals
p j (u) =
Z
Im(∂ j u · ū),
Rd
we can similarly compute
d
p j (u + εv)|ε=0 =
dε
Z
Im(v̄ · ∂ j u + ū · ∂ j v) = 2
Z
ω(∇ω p j (u), v) = −2
Im(∇ω p j (u) · v̄).
Rd
Z
Im(∂ j u · v̄),
Rd
Rd
Thus, ∇ω p j (u) = −∂ j u, so the associated Hamiltonian equation is the transport equation
∂t u = −∂ j u,
which has solution flows
u(t, x) = u(0, x − te j ),
where e j is the unit vector pointing in the positive x j -direction. As H is clearly conserved
by these flows, each p j is conserved by solutions of the Schrödinger equation. The corresponding symmetry for H is given by solutions of the p j -flows, which are translations in
the x j -direction. By taking each 1 ≤ j ≤ d, we obtain symmetry for all spatial translations.
Finally, for the normalized center-of-mass, which are time-dependent Hamiltonians, we
must first extend our “phase space” as in Exercise 1.42. Let D denote our informal “phase
space” for the linear Schrödinger equations, on which ω is defined. We define D̄ = R2 ×D,
and we define the following symplectic form on D̄:
ω̄((a, b, u), (a0 , b0 , u0 )) = ab0 − ba0 + ω(u, u0 ).
Again, as in Exercise 1.42, we extend H to a Hamiltonian on D̄:
H̄ ∈ C 1 (D̄ → R),
H̄(a, b, u) = H(u) + b.
A direct computation yields that
∇ω̄ H̄ = (1, 0, ∇ω H).
As a result, a curve t 7→ u(t) solves the linear Schrödinger equations, with initial value
u(0) = u0 ∈ D, if and only if for any b ∈ R, the curve t 7→ ub (t) = (t, b, u(t)) solves the
H̄-Hamilton equations, with initial value ub (0) = (0, b, u0 ).
We now define the normalized center-of-mass Hamiltonians on this extended phase
space D̄. Given 1 ≤ j ≤ d, we define the functions
Z
N j ∈ C 1 (D̄ → R),
N j (a, b, u) =
x j |u|2 dx − ap j (u).
Rd
To compute the symplectic gradient of N j , we first compute
Z
Z
d
N j (a + εa0 , b + εb0 , u + εv)|ε=0 = 2 Re
x j uv̄dx − a0 p j (u) − 2a
Im(∂ j u · v̄)
d
d
dε
ZR
ZR
= 2 Im
ix j uv̄dx − a0 p j (u) − 2a
Im(∂ j u · v̄).
Rd
Considering the definition of ω̄, then we see that
∇ω̄ N j (a, b, u) = (0, p j (u), −ix j u + a∂ j u).
Hence, the Hamilton equations associated with N j are
∂ s (a(s), b(s), u(s)) = (0, p j (u(s)), −ix j u(s) + a(s)∂ j u(s)).
Rd
30
ARICK SHAO
Now, suppose (a(s), b(s), u(s)) is a solution of the above equation. Then, for any s ∈ R,
d
d
H(u(s)) + p j (u(s))
H̄(a(s), b(s), u(s)) =
ds
ds
Z
d
=
Re ∇u(s) · ∇u(s) + p j (u(s))
d
ds
ZR
=
Re{∇[−ix j u(s) + a(s)∂ j u(s)]∇u(s)} + p j (u(s))
d
ZR
Z
1
=
Im u(s)∂ j u(s) + a(s)
∂ j |∇u(s)|2 + p j (u(s))
2
Rd
Rd
= −p j (u(s)) + p j (u(s))
= 0.
Thus, H̄ is conserved by the Hamilton flows of N j , and therefore by Noether’s theorem,
N j is conserved by the solutions of the linear Schrödinger equation. Finally, solving the
N j -Hamilton equation explicitly for a(s) and u(s), we see that
a(s) ≡ t0 ∈ R,
1
2
u(s) = e− 2 it0 s e−ix j s u0 (x + t0 se j ),
where e j ∈ Rd is the unit vector in the positive x j -direction. Since a(s) ≡ t0 corresponds to
the time variable, then the above curve s 7→ u(s) generates the Galilean symmetry indicated
in Exercise 2.5, in the special case v = −se j . 26 Combining all the above componentwise
Galilean symmetries for 1 ≤ j ≤ d yields the general Galilean symmetry.
2.48. Letting e0 = |∇u|2 /2, then
X X
iX
i
iX
∂ j ∆u · ∂ j u =
∂ j Re ∆u · ∂ j u − Re
|∆u|2 .
∂t e0 = Re
∂ j ∂t u · ∂ j u = Re
2
2
2
j
j
j
j
The second term on the right-hand side of course vanishes, while the first can be written as
the divergence of the vector field Re( 2i ∆u∇u). This is the desired local conservation law.
2.49. First, from the conservation of the pseudo-conformal energy, we have
k(x + it∇)u(t)kL2x (BR ) ≤ k(x + it∇)u(t)kL2x (Rd ) = kxu(0)kL2x (Rd ) .
As a result, by the conservation of mass, we have
k∇u(t)kL2x (BR ) ≤ |t|−1 [kxu(t)kL2x (BR ) + kxu(0)kL2x ]
≤ |t|−1 [Rku(t)kL2x (BR ) + kxu(0)kL2x ]
. hRi|t|−1 khxiu(0)kL2x .
2.50.
27
Let φ : Rd → R be a bump function such that
1 |x| ≤ 1,
φ(x) =
0 |x| ≥ 2.
Define
M(t) =
Z
φ2 (x/R)|u(t, x)|2 dx
Rd
26Note the correction at the beginning of Exercise 2.5.
27Thanks to Kyle Thompson for a technical observation.
! 21
,
t ∈ R.
SOLUTIONS
31
Differentiating the above and recalling the Schrödinger equations, we obtain
Z
φ2 (x/R) · i∆u(t, x) · u(t, x)dx
∂t M(t) . M(t)−1 Re
Rd
Z
|∇ x [φ(x/R)]||φ(x/R)u(t, x)||∇u(t, x)|dx,
. M(t)−1
Rd
where we have integrated by parts in the last step, noting that when the derivative hits u,
then the integrand is purely imaginary. Applying Hölder’s inequality yields
1
1
∂t M(t) . M(t)−1 k∇ x [φ(x/R)]kL∞x M(t)E 2 . R−1 E 2 .
Integrating the above results in the inequality
1
M(t) ≤ M(0) + Od (R−1 E 2 |t|).
Finally, by the definition of φ and M, we have for any t , 0 that
! 12
Z
|u(t, x)|2 dx ≤ M(t)
|x|≤R
1
≤ M(0) + Od (R−1 E 2 |t|)
! 12
Z
1
2
≤
|u(0, x)| dx + Od (R−1 E 2 |t|).
|x|≤2R
k,k
2.52. First, we expand the H -norm using the Plancherel theorem:
1
ke 2 it∆ f kHxk,k (Rd ) =
k
X
1
khxi j e 2 it∆ f kHxk− j (Rd )
j=0
'
k
X
1
2
khξik− j h∇i j (e 2 it|ξ| fˆ)kL2x (Rd )
j=0
.
X
1
2
khξia ∇b (e 2 it|ξ| fˆ)kL2x (Rd ) .
a+b≤k
Note that whenever a derivative hits the exponential factor, one picks up an extra factor of
ξ and t. Thus, applying the Leibniz rule and induction to the above yields
1
ke 2 it∆ f kHxk,k (Rd ) .
X
1
2
ktc hξia+c e 2 it|ξ| ∇b fˆkL2x (Rd ) . htik
j
k X
X
khξik− j ∇l fˆkL2x (Rd ) .
j=0 l=0
a+b+c≤k
Applying the Plancherel theorem, we have, by definition,
1
ke 2 it∆ f kHxk,k (Rd ) . htik
k
X
khxi j f kHxk− j (Rd ) . htik k f kHxk,k (Rd ) .
j=0
2.53. First, note we can assume ε > 0 is small, without loss of generality. We wish to
adapt the Morawetz-type argument using the smoothed function a(x) = hxi − εhxi1−ε .
From (2.37) and the definition of T 0 j , we have
Z
Z
∂t
∂ j a(x) · Im[u(t, x)∂ j u(t, x)] · dx =
∂ jk a(x) · Re(∂ j u(t, x)∂k u(t, x)) · dx
3
3
R
R
Z
1
|u(t, x)|2 ∆2 a(x) · dx
−
4 R3
= I1 + I2 .
32
ARICK SHAO
We now compute and estimate the derivatives of a.
∂ j a(x) = x j hxi−1 + (1 − ε)x j hxi−1−ε ,
!
xi x j
xi x j
∂i j a(x) = hxi [1 − ε(1 − ε )hxi ] δi j − 2 + hxi−3 [1 − ε(1 − ε2 )hxi−ε ] 2
|x|
|x|
−1
2
−ε
+ ε2 (1 − ε)hxi−1−ε δi j
!
xi x j
xi x j
= A1 · δi j − 2 + A2 · 2 + ε2 (1 − ε)hxi−1−ε δi j .
|x|
|x|
In particular, this implies the following estimates:
|∂ j a(x)| .ε 1,
|∂i j a(x)| .ε hxi−1 + hxi−3 + hxi−1−ε . hxi−1 .
Furthermore, we can compute
∆2 a(x) = −ε2 (1 + ε)(1 − ε)(2 − ε)hxi−3−ε
− hxi−5 {1 · 3 · (2 · 1 − 1) − ε(1 + ε)(3 + ε)[(2 + ε)(1 − ε) − 1]hxi−ε }
− hxi−7 [1 · 1 · 3 · 5 − ε(1 − ε)(1 + ε)(3 + ε)(5 + ε)hxi−ε ]
= −ε2 (1 + ε)(1 − ε)(2 − ε)hxi−3−ε − B1 − B2 .
From our computations for ∂2 a, we can now evaluate I1 . Note first that since ε is
sufficiently small, then the factors A1 and A2 are both everywhere nonnegative. As a result,
Z
Z
Z
A2 · |∂r u(t, x)|2 · dx + Cε
hxi−1−ε |∇u(t, x)|2 dx
A1 · |/
∇u(t, x)|2 · dx +
I1 =
R3
R3
R3
Z
≥ Cε
hxi−1−ε |∇u(t, x)|2 dx,
R3
where Cε > 0 is a constant depending on ε. Similarly, since B1 , B2 ≥ 0 everywhere,
Z
Z
Z
Dε
1
Dε
I2 =
hxi−3−ε |u(x, t)|2 dx +
(B1 + B2 )|u(x, t)|2 dx ≥
hxi−3−ε |u(x, t)|2 dx.
4 R3
4 R3
4 R3
Integrating our initial Morawetz-type identity over the time interval [−T, T ] yields
Z
Z TZ
t=T
∂ j a(x) · Im(u(t, x)∂ j u(t, x)) · dx
≥ Cε
hxi−1−ε |∇u(t, x)|2 dxdt
3
R3
−T
R
t=−T
Z Z
Dε T
hxi−3−ε |u(x, t)|2 dxdt,
+
4 −T R3
where we have applied the above observations and inequalities. Due to our estimates for a
along with mass conservation, we can apply the “momentum estimate” of Lemma A.10 to
bound the left-hand side by some constant times ku(0)k2Ḣ 1/2 . Finally, letting T % ∞ yields
Z Z
Z Z
−1−ε
2
hxi
|∇u(t, x)| dxdt +
hxi−3−ε |u(x, t)|2 dxdt .ε ku(0)k2 1 .
R
R3
R
R3
Ḣ 2
2.58. We first compute the symplectic gradient ∇ω H. Formally, we have
Z
d
1 d
H((u0 , u1 ) + ε(v0 , v1 ))|ε=0 =
(|∇u0 + ε∇v0 |2 + |u1 + εv1 |2 )|ε=0
dε
2 dε Rd
Z
=
(∇u0 · ∇v0 + u1 · v1 )
d
ZR
=
(u1 · v1 − ∆u0 · v0 ).
Rd
SOLUTIONS
33
In addition, letting (w0 , w1 ) = ∇ω H(u0 , u1 ), we have
Z
ω((w0 , w1 ), (v0 , v1 )) =
(w0 v1 − w1 v0 ),
Rd
so that by varying v0 and v1 , we have ∇ω H(u0 , u1 ) = (u1 , ∆u0 ). Consequently, the Hamilton
flow associated with H is ∂t (u0 , u1 ) = (u1 , ∆u0 ), or equivalently, the second-order system
∂2t u0 = ∂t u1 = ∆u0 ,
u0 |t=t0 = φ,
∂t u0 |t=t0 = u1 |t=t0 = ψ.
2.59. Recalling the identity (2.46) for the stress-energy tensor, we have
Z
Z
Z
Z
∂t
T 00 = −
∂t u · F,
∂i T i0 + Re
∂0 u · F = − Re
Rd
Rd
Rd
Rd
on any timeslice t = τ. Integrating the above with respect to the time over the interval [0, t]
and recalling the exact form of T 00 yields
Z tZ
|∂t u||F|dxdτ.
k∇u(t)k2L2 + k∂t u(t)k2L2 . k∇u0 k2L2 + ku1 k2L2 +
x
x
x
x
0
Rd
Taking a supremum over all t ≥ 0 and applying Hölder’s inequality, we have
k∇ukC2 0 L2 + k∂t u(t)kC2 ∞ L2 . k∇u0 k2L2 + ku1 k2L2 + k∂t ukLt∞ L2x kFkLt1 L2x .
x
t
x
t
x
x
Applying a weighted Cauchy inequality to the last term on the right-hand side completes
the proof of the energy estimate (2.28) in the case s = 1.
For general s ∈ R, note that the operator h∇i s−1 commutes with all derivatives, and
hence h∇i s−1 u also satisfies the wave equation
h∇i s−1 u = h∇i s−1 F.
Finally, applying the above estimate (the s = 1 case), we obtain
k∇ukC2 0 H s−1 + k∂t ukC2 0 H s−1 . k∇h∇i s−1 ukC2 0 L2 + k∂t h∇i s−1 ukC2 0 L2
t
x
t
x
t
x
t
x
. kh∇i s−1 FkLt1 L2x
. kFkLt1 Hxs−1 .
2.60. Elaboration: We assume our variation is compact, that is, X has compact support.
Using the given notations, we have
d
S (u s , g s )
0=
ds
s=0
Z
d
=
L(u s , g s )dg s ds R1+d
s=0
Z
XZ
∂L
d
∂L
d αβ
=
(u s , g s ) u s · dg s +
(u s , g s ) g s · dg s αβ
1+d ∂g
ds
ds
R1+d ∂u
s=0
α,β R
s=0
Z
d p
+
L(u s , g s )
| det g s |
ds
R1+d
s=0
Z
XZ
∂L
d
∂L
d
=
(u, g) u s | s=0 · dg +
(u, g) gαβ
s | s=0 · dg
αβ
1+d ∂g
ds
ds
R1+d ∂u
R
α,β
Z
d p
+
L(u, g)
| det g s |
ds
s=0
R1+d
= A + B + C.
34
ARICK SHAO
The term A corresponds to the Euler-Lagrange equation for the fixed-metric Lagrangian
T (u) = S (u, g). Since u is a critical point of T by assumption, A vanishes. To evaluate B
and C, we must evaluate derivatives of components of the metric. First,
d αβ βν d
= −gαµ gβν π .
g
g s = − gαµ
(g
)
µν
s µν s
s
ds
ds
s=0
s=0
Similarly, since | det g s | = − det g s (due to the Lorentzian signature), we can compute
1
1
d p
1
1
d
| det g s | = (− det g)− 2
(− det g s ) = (− det g) 2 · gµν πµν .
ds
2
ds
2
s=0
s=0
Combining the above observations, we obtain
0 = B+C
#
"
Z
1 µν
αµ βν ∂L
=
(u, g) + g L(u, g) πµν dg
−g g
∂gαβ
2
R1+d
Z
=−
gµα gνβ T αβ πµν dg
R1+d
Z
=−
T µν πµν dg.
R1+d
This completes the first part of the problem.
Since T αβ is symmetric in α and β, then
T αβ παβ = T αβ (∇α Xβ + ∇β Xα ) = 2T αβ ∇α Xβ .
As a result, by the (spacetime) divergence theorem,
Z
Z
0=
T αβ ∇α Xβ = −
R1+d
∇α T αβ · Xβ .
R1+d
Since this holds for arbitrary X (say, of compact support), then ∇α T αβ ≡ 0, i.e., T is
divergence-free. Finally, in the special case L(u, g) = gαβ ∂α u∂β u, we have
T αβ =
∂L
1
1
(u, g) − gαβ L(u, g) = ∂α u∂β u − gαβ gµν ∂µ u∂ν u.
∂gαβ
2
2
2.64. Let X denote the radial field
x
· ∇x .
|x|
We can compute the deformation tensor π with respect to X, here with respect to Cartesian
coordinates. First, since X is time-independent and has no time component, then
∂r =
π0β = πβ0 ≡ 0,
0 ≤ β ≤ 3.
Next, if i and j are spatial indices, i.e., 1 ≤ i, j ≤ 3, then
πi j = ∂i X j + ∂ j Xi = 2 ·
|x|2 δi j − xi x j
.
|x|3
Since T is divergence-free, then
1
∂α (T αβ Xβ ) = T i j πi j
2
1
= (∂i u∂ j u − δi j ∂α u∂α u)(|x|−1 δi j − |x|−3 xi x j )
2
|∇ x u|2
3
1
1
=
−
∂α u∂α u − (∂r u)2 +
∂α u∂α u
|x|
2|x|
|x|
2|x|
SOLUTIONS
35
1
|/
∇ x u|2
− ∂α u∂α u
|x|
|x|
|/
∇ x u|2
1
=
−
(|u|2 ).
|x|
2|x|
Next, fix a cutoff function η on R supported on [−1, 1], fix T 0 > 0, and define the
rescaling ηT0 (t) = η(t/T 0 ). On one hand, by the spacetime divergence theorem,
Z Z
Z Z
αβ
ηT0 (t) · ∂α (T Xβ ) · dxdt =
∂α [ηT0 (t) · T αβ Xβ |(t,x) ]dxdt
R R3
R R3
Z Z
−
∂t ηT0 (t) · T 0β Xβ · dxdt
=
R
= 0 − I1 .
R3
Note in particular that due to the cutoff function ηT0 and the rapid decay of u in the spatial
directions, the divergence theorem yields no boundary terms. Moreover, since X has unit
length everywhere, and since |T αβ | . |D x,t u|2 , then
Z T0 Z
Z T0 Z
1
|Dt,x u|2 dxdt . E,
|I1 | . k∂t ηT0 kLt∞
|Dt,x u|2 dxdt .η
3
3
T
0
−T 0 R
−T 0 R
where
1
1
ku(0)kḢx1 + k∂t u(0)kL2x ,
2
2
and where we have applied the standard energy conservation for the wave equation.
On the other hand, we also have that from our expansion for ∂α (T αβ ) that
Z Z
Z Z
1
ηT0 (t)[|/
∇ x u|2 + ∂2t (|u|2 ) − ∆(|u|2 )]dxdt
ηT0 (t)∂α (T αβ Xβ )|(t,x) dxdt =
R R3
R R3 |x|
= I2 + I3 + I4 .
E=
The term I2 , we can simply leave alone. For I4 , recalling that the distribution −(4π|x|)−1 is
the fundamental solution of ∆ (with respect to the origin of R3 ), then we can compute 28
Z
I4 = 4π ηT0 (t)|u(0, t)|2 dt.
R
Lastly, for I3 , we integrate by parts once and apply Hölder’s inequality:
! 21
Z T0
Z T0 Z
Z
|∂t (|u|2 )|
1
|u(x, t)|2
|I3 | . k∂t ηT0 kLt∞
dxdt .η
k∂t u(t)kL2x
dx dt.
T 0 −T0
|x|2
|x|2
−T 0 R3
R3
Applying Hardy’s inequality (Lemma A.2) to the spatial integral, we have
Z T0
1
k∂t u(t)kL2x k∇ x u(t)kL2x dt . E,
|I3 | .
T 0 −T0
where we have again applied the conservation of energy.
Combining all the above yields our desired inequality
Z Z
Z
|/
∇ x u(t, x)|2
ηT0 (t)
dxdt +
ηT0 (t)|u(t, x)|2 ' I2 + I4 . E.
|x|2
R R3
R
Moreover, note that the above inequality holds independently of T 0 . Thus, letting T 0 % ∞
and applying the dominated convergence theorem yields the desired Morawetz estimate.
28Of course, one can compute I explicitly, by replacing the spatial integral over R3 by the same integral over
4
R3 minus a ball of radius ε about the origin and then letting ε & 0.
36
ARICK SHAO
2.69. First, we can directly compute
1
∂α T αβ = F∂β u + ∂α u · ∂α ∂β u − ∂β (∂α u · ∂α u) = F∂β u.
2
Consider the vector field 29
X = e2tx j ∂ j ,
By the divergence theorem,
Z
Z
0=
∂α (T αβ Xβ ) =
Rd
Rd
Xk = δ jk e2txk .
F∂β u · Xβ +
Z
Rd
T αβ ∂α Xβ = −I1 + I2 .
For I1 , we can expand
I1 = −
Z
Rd
e2tx j F∂ j u = −
Z
Rd
etx j F∂ j (etx j u) + t
Z
Fue2tx j .
Rd
Next, since ∂α Xβ = δα j δβ j 2te2tx j , then
Z
T j j e2tx j
I2 = 2t
Rd
Z
Z
j 2 2tx j
= 2t
|∂ u| e − t
∂i u∂i ue2tx j
Rd
Rd
Z
Z
Z
Z
= 2t
etx j ∂ j u∂ j (etx j u) − 2t2
∂ j u · u · e2tx j + t
uFe2tx j + 2t2
u · ∂ j u · e2tx j
d
d
d
d
R
R
ZR
Z
Z R
tx j
2
2
tx j
tx j
2tx j
= 2t
|∂ j (e u)| − 2t
e u · ∂ j (e u) + t
uFe .
Rd
Rd
Rd
Since the second term on the right vanishes (by the fundamental theorem of calculus), and
since I1 = I2 by our previous calculations, then
Z
Z
2t
|∂ j (etx j u)|2 = −
etx j F∂ j (etx j u).
Rd
Rd
Finally, applying Hölder’s inequality to the above yields the Carleman inequality
k∂ j (etx j u)kL2 ≤
1 tx j
ke FkL2 .
2|t|
Suppose now that ∆u = O(|u|). Since u is compactly supported, then
"Z Z
# 21
Z Z xj
1 tx j 2
ts
tx j
tx j
ts
2
ke ukL2 =
∂ j (e u) · e u · dsdx ≤ ke ukL2
|∂ j (e u)| dxds .
2
Rd −∞
R Rd
If R is chosen so that u is supported entirely in the region {|x j | ≤ R}, then
ketx j ukL2 ≤ 2Rk∂ j (etx j u)kL2 ≤
R tx j
ke ukL2 ,
|t|
where we applied the Carleman inequality in the last step. Taking t to be sufficiently large
forces ketx j ukL2 = 0, which implies u ≡ 0 and proves the unique continuation property.
29Here, t is simply a nonzero constant.
SOLUTIONS
37
2.70. Correction: The correct identity we wish to show is
= kvkHtb Hxs .
kukX s,b
τ=h(ξ)
Let F x and Ft denote Fourier transforms in space and time, respectively. Since
u(t) = U(t)v(t) = etL v(t),
F x u(t) = eith(ξ) F x v(t),
then we have
kukX s,b
τ=h(ξ)
= khξi s hτ − h(ξ)ib · Ft [eith(ξ) F x v(t)]kLτ2 Lξ2
= khξi s hτ − h(ξ)ib · Ft F x [v(τ − h(ξ), ξ)]kLτ2 Lξ2
= khξi s hτib · Ft F x [v(τ, ξ)]kLτ2 Lξ2
= kvkHtb Hxs ,
where we applied the Plancherel theorem in the last step.
2.75. 30 We first prove the analogous estimate for solutions of the linear Schrödinger
equation. More specifically, we show that if u0 , v0 ∈ S x (Rd ), and if û0 and v̂0 are supported
in the Fourier domains |ξ| ≤ M and |ξ| ≥ N, respectively, then
keit∆ u0 eit∆ v0 kLt2 L2x .d
M
d−1
2
1
N2
ku0 kL2x kv0 kL2x .
Again, if d = 1, we also require that N > 2M.
First, suppose d ≥ 2 and N .d M. Then, applying the Gagliardo-Nirenberg inequality
(Proposition A.3) and the Strichartz inequality, we obtain, as desired,
keit∆ u0 eit∆ v0 kLt2 L2x .d keit∆ u0 kLt4 L2dx keit∆ v0 k
.d k|∇|
.d k|∇|
.M
.d
d−2
2
M
d−2
2
d−2
2
eit∆ u0 k
2d
Lt4 L xd−1
2d
Lt4 L xd−1
keit∆ v0 k
2d
Lt4 L xd−1
u0 kL2x kv0 kL2x
ku0 kL2x kv0 kL2x
d−1
2
ku0 kL2x kv0 kL2x .
1
N2
Next, we consider any arbitrary dimension d, but with N d M (in the case d = 1, we
need only assume that N > 2M). By duality, it suffices to prove
Z Z
d−1
M 2
I = eit∆ u0 (x)eit∆ v0 (x)F(t, x)dxdt .d
ku0 kL2x kv0 kL2x kFkLt2 L2x .
1
R Rd
N2
By Parseval’s identity, and recalling that
Ft,x (eit∆ u0 )(τ, ξ) = δ(τ − |ξ|2 )û0 (ξ),
Ft,x (eit∆ v0 )(τ, ξ) = δ(τ − |ξ|2 )v̂0 (ξ),
in the distributional sense, we can expand I as follows:
Z Z
I ' Ft,x (eit∆ u0 eit∆ v0 )(τ, ξ)F̃(τ, ξ)dξdτ
d
ZR ZR Z
2
2
'
δ(τ1 − |ξ1 | )û0 (ξ1 )δ(τ2 − |ξ2 | )v̂0 (ξ2 )dξ1 dτ1 F̃(τ, ξ)dξdτ
R Rd (τ1 ,ξ1 )+(τ2 ,ξ2 )=(τ,ξ)
30Much of the solution was obtained from [2].
38
ARICK SHAO
Z Z
û0 (ξ1 )v̂0 (ξ2 )F̃(|ξ1 |2 + |ξ2 |2 , ξ1 + ξ2 )dξ1 dξ2 .
' Rd
Rd
Applying Hölder’s inequality and recalling the supports of û0 and v̂0 , then
21
Z
"Z
#2
v̂0 (ξ2 )F̃(|ξ1 |2 + |ξ2 |2 , ξ1 + ξ2 )dξ2 dξ1
I . ku0 kL2x
|ξ1 |≤M Rd
"Z
# 12
Z
2
2
2
. ku0 kL2x kv0 kL2x
|F̃(|ξ1 | + |ξ2 | , ξ1 + ξ2 )| dξ2 dξ1 .
|ξ1 |≤M
|ξ2 |≥N
Given 1 ≤ i ≤ d, we define the domain
1
Di = {(ξ1 , ξ2 ) ∈ Rd × Rd | |ξ1 | ≤ M, |ξ2 | ≥ N, |ξ2i | ≥ Nd− 2 },
where ξ1i is the i-th component of ξ1 . Note that
d
[
Di = {(ξ1 , ξ2 ) ∈ Rd × Rd | |ξ1 | ≥ M, |ξ2 | ≥ N}.
i=1
Consider the change of variables
s = ξ1 + ξ2 ,
r = |ξ1 |2 + |ξ2 |2 ,
ξ10 = ξ10 ,
on Di , where ξ10 ∈ Rd−1 represents ξ1 ∈ Rd but with the i-th component ξ1i omitted. An
explicit calculation yields the following value for the corresponding Jacobian:
∂(r, ξ10 , s) J= i 0
= 2|ξ1i ± ξ2i |,
∂(ξ , ξ , ξ2 ) 1 1
Here, the sign in “±” depends on the dimension d. By our assumption N d M (or N > 2M
when d = 1) and from our definition of Di , we have that J & N on Di . Integrating now
over Di and applying this change of variables, we have
Z
Z
Z Z
2
2
2
0
|F̃(|ξ1 | + |ξ2 | , ξ1 + ξ2 )| dξ1 dξ2 .
dξ1
|F̃(r, s)|2 kJ −1 kL∞ (Di ) dsdr
|ξ10 |≤M
Di
Rd
R
.d M d−1 N −1 kFk2L2 L2 .
t
x
As a result, combining all the above, we obtain
d Z
12
d−1
X
M 2
ku0 kL2x kv0 kL2x kFkLt2 L2x .
I . ku0 kL2x kv0 kL2x
|F̃(|ξ1 |2 + |ξ2 |2 , ξ1 + ξ2 )|2 dξ2 dξ1 .d
1
N2
i=1 Di
This completes the proof of the case N d M. As a result, we have proved
keit∆ u0 eit∆ v0 kLt2 L2x .d
M
d−1
2
1
N2
ku0 kL2x kv0 kL2x ,
with u0 and v0 as before. It remains to convert the above into an X s,b -type estimate.
Let u, v ∈ St,x (R × Rd ) satisfy the hypotheses in the problem statement. Given any
σ, τ ∈ R, we define the following functions on Rd :
Z
Z
−d
2
ix·ξ
−d
fσ (x) = (2π)
ũ(|ξ| + σ, ξ)e dξ,
gτ (x) = (2π)
ṽ(|ξ|2 + τ, ξ)eix·ξ dξ.
Rd
Rd
SOLUTIONS
39
Note in particular that fˆσ and ĝτ are supported in the Fourier domains |ξ| ≤ M and |ξ| ≥ N,
respectively, for any σ and τ. From the proof of Lemma 2.9, we have the identities
Z
Z
1
1
itσ it∆
u(t) =
e e fσ dσ,
v(t) =
eitτ eit∆ gτ dτ.
2π R
2π R
Using the above identities, we obtain
Z Z
eitσ eit∆ fσ eitτ eit∆ gτ dσdτ
kuvkLt2 L2x . R R
Lt2 L2x
Z Z
.
keit∆ fσ eit∆ gτ kLt2 L2x dσdτ
R
.d
M
R
d−1
2
Z
Z
k fσ k dσ kgτ kL2x dτ,
1
R
N2 R
where in the last step, we applied the free Schrödinger estimate established above. Finally,
applying Hölder’s inequality as in the proof of Lemma 2.9, we have
# 21 "Z
# 12
d−1 "Z
d−1
M 2
M 2
2b
2
2b
2
kuvkLt2 L2x .d,b
kukX 0,b kvkX 0,b .
hσi
k
f
k
hτi
kg
k
.
dσ
dτ
2
2
σ
τ
1
1
Lx
Lx
τ=|ξ|2
τ=|ξ|2
R
R
N2
N2
L2x
Chapter 3: Semilinar Dispersive Equations
3.1. First, for the NLS, consider the symplectic form ω and the Hamiltonian H, given by
!
Z
Z
1
2
ω(u, v) = −2
Im(uv̄),
H(u) =
|∇u|2 +
µ|u| p+1 ,
p+1
Rd
Rd 2
where u and v are in the appropriate spaces. Taking a (directional) derivative of H yields
#
Z "
d
1
2
d
2
p+1 H(u + εv)|ε=0 =
|∇(u + εv)| +
µ|u + εv|
dε
dε Rd 2
p+1
ε=0
Z
p−1
=
[Re(∇u · ∇v̄) + 2µ|u| Re(uv̄)]
Rd
Z
(−i∆u + 2iµ|u| p−1 u)v̄
= Im
Rd
!
1
p−1
= ω i∆u − iµ|u| u, v .
2
As a result, the symplectic gradient of H is
1
i∆u − iµ|u| p−1 u,
2
and hence the Hamiltonian evolution equation is
∇ω H(u) =
1
1
i∆u − iµ|u| p−1 u,
i∂t u + ∆u = µ|u| p−1 u.
2
2
Similarly, for the NLW, we define ω and H, also on appropriate spaces, by
Z
ω((u0 , u1 ), (v0 , v1 )) =
(u0 v1 − v0 u1 ),
Rd
!
Z
1
1 2
1
2
p+1
H(u0 , u1 ) =
|∇u0 | + |u1 | +
µ|u0 |
.
2
p+1
Rd 2
∂t u =
40
ARICK SHAO
Again, taking a directional derivative yields
Z
d
H(u0 + εv0 , u1 + εv1 )|ε=0 =
[∇u0 · ∇v0 + u1 v1 + µ|u0 | p−1 (u0 v0 )]
dε
Rd
Z
=
[u1 v1 − (∆u0 − µ|u0 | p−1 u0 )v0 ]
Rd
= ω((u1 , ∆u0 − µ|u0 | p−1 u0 ), (v0 , v1 ))
Thus, the symplectic gradient of H and the associated Hamiltonian evolution equation are
∇ω H(u0 , u1 ) = (u1 , ∆u0 − µ|u0 | p−1 u0 ),
∂t u0 = u1 , ∂t u1 = ∆u0 − µ|u0 | p−1 u0 .
Combining the above, we obtain the nonlinear wave equation
u0 = −∂2t u0 + ∆u0 = −∂t u1 + ∆u0 = µ|u0 | p−1 u0 .
3.2. Let u and v be defined as in the problem statement. For convenience, we also define
t − x
d+1
, x1 , . . . , xd ,
E = e−i(t+xd+1 ) .
z = (t, x) = (t, x1 , . . . , xd+1 ),
z0 =
2
We can then compute
1
∂t v(z) = −iEu(z0 ) + E∂t u(z0 ),
2
1
2
0
∂t v(z) = −Eu(z ) − iE∂t u(z0 ) + E∂2t u(z0 ),
4
1
0
0
∂ xd+1 v(z) = −iEu(z ) − E∂t u(z ),
2
1
2
0
∂ xd+1 v(z) = −Eu(z ) + iE∂t u(z0 ) + E∂2t u(z0 ).
4
Furthermore, we define the symbols
∆d =
d
X
∂2xk ,
∆d+1 =
k=1
d+1
X
∂2xk .
k=1
Combining all the above, we compute
−∂2t v(z) + ∆d+1 v(z) = −∂2t v(z) + E∆d u(z0 ) + ∂2xd+1 v(z) = 2iE∂t u(z0 ) + E∆d u(z0 ).
Since u satisfies the NLS, then
−∂2t v(z) + ∆d+1 v(z) = 2µE|u(z0 )| p−1 u(z0 ) = 2µ|Eu(z0 )| p−1 Eu(z0 ) = 2µ|v(z)| p−1 v(z).
Correction: If u satisfies NLS, then v satisfies NLW, with an extra factor of 2 multiplied
to the power nonlinearity. This comes from the factor of 1/2 on the Laplacian in the NLS.
3.5. 31 Correction: The solutions uv,λ in (3.21) of the nonperiodic focusing NLS, being
Galilean transforms of rescaled soliton solutions, should be
x − vt t|v|2
tτ
2
.
uv,λ = λ− p−1 eix·v e−i 2 +i λ2 Q
λ
Throughout, we will always fix the constant τ to be 1. 32
Correction: The statement we will actually show is the following. Suppose s < 0 or
s < sc , and let 0 < δ ε . 1. Then there exist solutions u and u0 to (3.1) such that:
• At time 0, both u and u0 have H s -norm comparable to ε.
31Part of the solution was inspired by [1].
32Since the soliton Q itself depends on τ, we fix τ a priori.
SOLUTIONS
41
• At time 0, the H s -separation between u0 and u is comparable to δ.
• At some later time t . ε p , for some positive power p depending on s, the H s separation between u0 and u is comparable to ε.
This shows that the solution map for (3.1) is not uniformly continuous in the H s -norm. In
particular, the requirement δ . ε is mandatory, since by the triangle inequality, if solutions
u and u0 have H s -norm comparable to ε at time 0, then
ku0 (0) − u(0)kHxs (Rd ) ≤ ku0 (0)kHxs (Rd ) + ku0 (0)kHxs (Rd ) . ε.
We begin by noting, for arbitrary v ∈ Rd and λ > 0, the identity
Z
x − vt t|v|2
t
2
eix·(v−ξ) Q
dx
ûv,λ (ξ) = λ− p−1 e−i 2 +i λ2
λ
Rd
Z
t|v|2
t
2
ei(λx+vt)·(v−ξ) Q (x) dx
= λd− p−1 e−i 2 +i λ2
= λd−
2
p−1
it
e
|v|2
2
Rd
−v·ξ+λ−2
Q̂(λξ − λv),
where ûv,λ denotes the spatial Fourier transform of uv,λ .
First, we consider the case
d
2
sc = −
> 0,
0 ≤ s < sc .
2 p−1
For conciseness, we write uλ in the place of u0,λ , for any λ > 0. Note that
Z
4
2d− p−1
2
kuλ (t)kḢ s (Rd ) = λ
|ξ|2s |Q̂(λξ)|2 dξ
x
Rd
Z
4
|ξ|2s |Q̂(ξ)|2 dξ
= λd− p−1 −2s
=λ
2(sc −s)
Rd
2
kQkḢ s (Rd ) .
x
In addition, fix λ0 > 0, and define γ = λ0 /λ.
Next, we compute the H s -separation at time t:
Z
2
it
it
2
2
kuλ0 (t) − uλ (t)k2Ḣ s (Rd ) =
|ξ|2s (λ0 )d− p−1 e (λ0 )2 Q̂(λ0 ξ) − λd− p−1 e λ2 Q̂(λξ) dξ
x
Rd
0 2(sc −s)
= (λ )
kQk2Ĥ s (Rd ) + λ2(sc −s) kQk2Ĥ s (Rd )
x
x
Z
2
2
d− p−1
¯
0 d− p−1
it[(λ0 )−2 −λ−2 ]
− 2(λ )
λ
Re e
|ξ|2s Q̂(λ0 ξ)Q̂(λξ)dξ
d
R
h
i
= λ2(sc −s) 1 + γ2(sc −s) kQk2Ḣ s (Rd )
x
Z
2
0 2
d
d
it λ 2−(λ0 )2
¯
0 sc + 2 sc − 2 +2s
λ
(λ
)
− 2(λ )
λ
Re e
|ξ|2s Q̂(γξ)Q̂(ξ)dξ
Rd
h
i
γ−2 −1
d
= λ2(sc −s) 1 + γ2(sc −s) kQk2Ḣ s (Rd ) − 2γ sc + 2 Re(eit λ2 · Iγ ) ,
x
where Iγ is the integral
Iγ =
Z
¯
|ξ|2s Q̂(γξ)Q̂(ξ)dξ.
Rd
If t = 0, then by the dominated convergence theorem,
lim Iγ = kQk2Ḣ s (Rd ) ,
γ→1
x
lim kuλ0 (0) − uλ (0)kḢxs (Rd ) = 0.
γ→1
42
ARICK SHAO
Thus, by choosing γ to be sufficiently close to 1, we have
kuλ0 (0) − uλ (0)kḢxs (Rd ) ≤ c · λ sc −s kQkḢxs (Rd ) = ckuλ (0)kḢxs ' ckuλ0 (0)kḢxs ,
for some small constant c 1 depending on γ, but independent of λ.
Furthermore, since γ is near 1, then Iγ is almost real-valued, so by taking some
t . λ2 · |γ−2 − 1|−1 ,
−2
−2
i.e., t to be λ2 times a large but fixed constant, the quantity eitλ (γ −1) Iγ becomes purely
imaginary, and it follows for this choice of t that
h
i
kuλ0 (t) − uλ (t)k2Ḣ s (Rd ) = λ2(sc −s) 1 + γ2(sc −s) kQk2Ḣ s (Rd ) ' λ2(sc −s) kQk2Ḣ s (Rd ) .
x
x
x
Now, since s ≥ 0, we have that
k f k2Hxs (Rd ) ' k f k2Ḣ s (Rd ) + k f k2Ḣ 0 (Rd ) .
x
x
As a result, we can apply the above computations both for s and for s = 0. Thus, given ε
and δ as in the problem statement, we can choose λ such that
kuλ0 (0)kHxs (Rd ) ' kuλ (0)kHxs (Rd ) ' ε,
kuλ0 (0) − uλ (0)kHxs (Rd ) ' δ ε,
kuλ0 (t) − uλ (t)kHxs (Rd ) ' ε.
This completes the proof in the case sc > 0 and 0 ≤ s < sc .
It remains to consider the case s < 0. For this, we define the shorthand uv for uv,1 for
any v ∈ Rd . We begin by computing the H s -norm at t = 0:
Z
Z
2
2 s
2
kuv (0)kHxs (Rd ) =
(1 + |ξ| ) |Q̂(ξ − v)| dξ =
(1 + |ξ + v|2 ) s |Q̂(ξ)|2 dξ.
Rd
Rd
In addition, fix v = (1 + β)v ∈ R , for sufficiently small β > 0. Moreover, we let
0
3
ε = kuv (0)kHxs (Rd ) ' kuv0 (0)kHxs (Rd ) .
To obtain a possible range of values of ε, we take |v| ≥ 1, and we write
Z
Z
kuv (0)k2Hxs (Rd ) =
(1 + |ξ + v|2 ) s |Q̂(ξ)|2 dξ +
(1 + |ξ + v|2 ) s |Q̂(ξ)|2 dξ = I1 + I2 .
|ξ|≥ |v|2
|ξ|≤ |v|2
Note first that
Z
I1 ' |v|
2s
|Q̂(ξ)|2 dξ ' |v|2s .
Rd
Next, since Q is rapidly decreasing, so is Q̂, hence for any α > 0,
Z
(1 + |ξ + v|2 ) s (1 + |ξ|2 )−α dξ . |v|−2α+d .
I2 .
|ξ|≥ |v|2
As long as the power α > 0 is chosen to be large enough, we obtain
kuv (0)k2Hxs (Rd ) ' |v|2s .
Since s < 0, it follows that
lim kuv (0)kHxs (Rd ) → 0.
v→∞
Thus, by continuity, we can choose ε to be any non-large constant, as desired.
For the time separation, we compute
0 2
2
2
Z
it |v | −v0 ·ξ+1
it |v| −v·ξ+1
kuv0 (t) − uv (t)k2Hxs (Rd ) =
(1 + |ξ|2 ) s e 2
Q̂(ξ − v0 ) − e 2
Q̂(ξ − v) dξ
Rd
SOLUTIONS
43
Z
(1 + |ξ|2 ) s |Q̂(ξ − v0 )|2 dξ +
(1 + |ξ|2 ) s |Q̂(ξ − v)|2 dξ
Rd
Rd
( |v0 |2 −|v|2
)
Z
it
−βv·ξ
2 s
0 ¯
2
−2
(1 + |ξ| ) Re e
Q̂(ξ − v )Q̂(ξ − v) dξ
Rd
Z
Z
=
(1 + |ξ + v0 |2 ) s |Q̂(ξ)|2 dξ +
(1 + |ξ + v|2 ) s |Q̂(ξ)|2 dξ
d
d
R
)
( R(β2 +2β)|v|2
Z
−βv·(ξ+v)
it
¯
2 s
2
Q̂(ξ − βv)Q̂(ξ)
−2
dξ
(1 + |ξ + v| ) Re e
Rd
Z
Z
=
(1 + |ξ + v0 |2 ) s |Q̂(ξ)|2 dξ +
(1 + |ξ + v|2 ) s |Q̂(ξ)|2 dξ
Rd
Rd
Z
2 2
¯
it β 2|v|
(1 + |ξ + v|2 ) s eitβv·ξ Q̂(ξ − βv)Q̂(ξ)dξ
− 2 Re e
Rd
Z
Z
(1 + |ξ + v0 |2 ) s |Q̂(ξ)|2 dξ +
(1 + |ξ + v|2 ) s |Q̂(ξ)|2 dξ − 2Jt,β ,
=
=
Z
Rd
Rd
where
Jt,β = Re eit
β2 |v|2
2
Z
Rd
¯
(1 + |ξ + v|2 ) s eitβv·ξ Q̂(ξ − βv)Q̂(ξ)dξ.
By the dominated convergence theorem, we have
Z
(1 + |ξ + v|2 ) s |Q̂(ξ)|2 dξ,
lim kuv0 (0) − uv (0)kHxs (Rd ) = 0.
lim J0,β =
β&0
β&0
Rd
Thus, by choosing β to be sufficiently small, we obtain for some small c 1 that
Z
2
2
kuv0 (0) − uv (0)kHxs (Rd ) ≤ c
(1 + |ξ + v|2 ) s |Q̂(ξ)|2 dξ = c2 kuv (0)k2Hxs (Rd ) = c2 ε2 .
Rd
Next, take t = β−2 |v|−1 , i.e., t a large constant times |v|−1 ' ε−1/s . Choosing a component
1 ≤ m ≤ d of v such that |vm | ' |v|, then we can bound
Z
1
¯ ·
iβ−1 v|v|−1 ·ξ
∂
e
dξ
|Jt,β | . (1 + |ξ + v|2 ) s Q̂(ξ − βv)Q̂(ξ)
−1 v |v|−1 m
d
β
m
ZR
¯
.β
|∂m [(1 + |ξ + v|2 ) s Q̂(ξ − βv)Q̂(ξ)]|dξ,
Rd
where we integrated by parts in the last step. Recalling that Q̂ is rapidly decreasing, we
can, using the same techniques as before, derive the bound
|Jt,β | . β|v|2s ' βε.
As a result, with β sufficiently small, and with this choice of t ' ε−1/s , we have
kuv0 (t) − uv (t)k2Hxs (Rd ) ' ε2 ,
kuv0 (t) − uv (t)kHxs (Rd ) ' ε.
Recall that, with δ = cε, we also had
kuv0 (0)kHxs (Rd ) ' kuv (0)kHxs (Rd ) ' ε,
kuv0 (0) − uv (0)kHxs (Rd ) ' δ ε.
This completes the proof for the case s < 0.
44
ARICK SHAO
3.6. In general, (3.2) has the conjugation invariance property: if u is a classical solution of
(3.2), then its conjugate ū also solves (3.2). To see this, we simply compute:
ū + ∆ū − µ|ū| p−1 ū = u + ∆u − µ|u| p−1 u ≡ 0.
Correction: We will prove the following: if u is a classical solution of (3.2), and if both
u(t0 ) and ∂t u(t0 ) are real-valued, then u is everywhere real-valued. Note that the additional
condition on ∂t u(t0 ) is necessary, since if u(t0 ) and ∂t u(t0 ) are purely real and imaginary,
respectively, then u cannot be everywhere real at a time near t0 .
If u is as above, then ū is also a classical solution of (3.2), and at t0 , it satisfies
ū(t0 ) = u(t0 ),
∂t ū(t0 ) = ∂t u(t0 ) = ∂t u(t0 ),
since both u(t0 ) and ∂t u(t0 ) are real-valued. Thus, by uniqueness (Proposition 3.3), it follows that u and ū are everywhere equal, and hence u is everywhere real-valued.
3.7. Let R ∈ SO(d, R) denote an arbitrary spatial rotation. Suppose u and v are classical
solutions to (3.1) and (3.2), respectively. By spatial rotation symmetry, the functions
uR , vR : I × Rd → C,
uR (t, x) = u(t, Rx),
vR (t, x) = v(t, Rx)
are also classical solutions to (3.1) and (3.2), respectively.
Now, suppose u(t0 ) is spherically symmetric. Then, uR solves (3.1), and uR (t0 ) = u(t0 ).
By uniqueness (see Proposition 3.2), it follows that uR = u everywhere. Since this is true
for any rotation R, then u is spherically symmetric.
Likewise, if v[t0 ] = (v(t0 ), ∂t v(t0 )) is spherically symmetric, then vR solves (3.2), and
vR [t0 ] = v[t0 ]. By uniqueness (Proposition 3.3), vR = v everywhere. By varying over all
rotations R, it follows that v is spherically symmetric.
3.9. Correction: In this problem, we are considering the focusing NLW.
Fix t0 > 0, and consider the solution to the focusing NLW in (3.6):
"
# 1
−2
2(p + 1) p−1
p−1
u(t, x) = c p (t0 − t) ,
cp =
.
(p − 1)2
This is a smooth solution that blows up at time t0 . Let ϕ : Rd → [0, 1] be a smooth and compactly supported cutoff function that is identically 1 on the ball about the origin of radius
R, with R t0 . Consider the compactly supported initial data (u0 , u1 ) = (ϕu(0), ϕ∂t u(0)),
which we impose at time t = 0. Solving the focusing NLW with this data yields a classical
solution v. 33 By uniqueness and finite speed of propagation (Proposition 3.3), it follows
that u and v must coincide on a cylinder C = {(t, x) | 0 ≤ t < t0 , |x| < r}, for some r > 0.
Since u blows up at t = t0 on C, then v blows up at t = t0 on C as well.
3.10. Since u is a strong H s -solution to (3.1), with data u(t0 ) = u0 , we have, by definition,
Z t
1
1
0
u ∈ Ct,loc
H xs (I × Rd ),
u(t) = e 2 i(t−t0 )∆ u0 − iµ
e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]dτ
t0
for any t ∈ I. We can restate the above in terms of t1 rather than t0 : 34
Z t
1
1
1
u(t) = e 2 i(t−t1 )∆ e 2 i(t1 −t0 )∆ u0 − iµ
e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]dτ
t1
33Here, we must apply existence theorems for classical solutions of the NLW.
34Note that in the last term below, we implicitly used that the strong solution u is continuous in time in order
to factor the linear propagator ei(t−t1 )∆/2 out of the time integral.
SOLUTIONS
t1
( Z
1
− e 2 i(t−t1 )∆ iµ
45
)
1
e 2 i(t1 −τ)∆ [|u(τ)| p−1 u(τ)]dτ .
t0
Since u1 = u(t1 ), and since u is a strong H s -solution to (3.1) with data u0 , then
Z t1
1
1
u1 = u 2 i(t1 −t0 )∆ u0 − iµ
e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]dτ.
t0
Consequently, we obtain, as desired
1
u(t) = e 2 i(t−t1 )∆ u1 − iµ
t
Z
1
e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]dτ,
t1
and it follows that u is a strong H -solution to (3.1), with data u(t1 ) = u1 .
Next, since conjugation preserves the H s -norm, the function
s
t 7→ ũ(t) = u(−t),
−t ∈ I
s
is also a continuous map into H . Moreover, by definition, for any such t, we have
Z −t
1
1
ũ(t) = e 2 i(−t−t0 )∆ u(t0 ) + iµ
e 2 i(−t−τ)∆ [|u(τ)| p−1 u(τ)]dτ.
t0
Since the linear Schrödinger equation is conjugation-invariant, then
Z −t
1
1
ũ(t) = e 2 i(t+t0 )∆ u(t0 ) + iµ
e 2 i(t+τ)∆ [|u(τ)| p−1 u(τ)]dτ
t0
=e
1
2 i(t+t0 )∆
=e
1
2 i(t+t0 )∆
Z
t
1
e 2 i(t−τ)∆ [|ū(−τ)| p−1 ū(−τ)]dτ
[ũ(−t0 )] − iµ
−t0
Z
t
ū0 − iµ
e
1
2 i(t−τ)∆
[|ũ(τ)| p−1 ũ(τ)]dτ.
−t0
The above equation shows that ũ is a strong H s -solution to (3.1), with data ũ(−t0 ) = ū0 .
Finally, let v be a strong H s -solution to (3.2) on an interval I, with initial datum
v[t0 ] = (v(t0 ), ∂t v(t0 )) = (v0 , v00 ),
t0 ∈ I.
As before, fix another time t1 ∈ I, and let
v[t1 ] = (v(t1 ), ∂t v(t1 )) = (v1 , v01 ).
For simplicity, we write
√
√
sin(s −∆)
L(s)( f, g) = cos(s −∆) f +
g,
√
−∆
L(s)( f, g) = (L(s)( f, g), ∂ s [L(s)( f, g)]).
representing the linear propagator for the wave equation, written as a first-order system.
Applying the semigroup property for L, the proof proceeds like in the case of the NLS:
Z t
(v(t), ∂t v(t)) = L(t − t0 )(v0 , v00 ) − µ
L(t − τ)(0, |v(τ)| p−1 v(τ))dτ
t0
Z t
L(t − τ)(0, |v(τ)| p−1 v(τ)])dτ
= L(t − t1 )[L(t1 − t0 )(v0 , v00 )] − µ
t1
( Z t1
)
L(t − t1 ) iµ
L(t1 − τ)(0, |v(τ)| p−1 v(τ))dτ
t0
= L(t −
t1 )(v1 , v01 )
−µ
Z
t1
t
L(t − τ)(0, |v(τ)| p−1 v(τ)])dτ.
46
ARICK SHAO
Thus, v is a strong H s -solution to (3.2), with data v[t1 ] = (v1 , v01 ).
3.12. Suppose u is a weak H s -solution to (3.1) with data u(t0 ) = u0 , where s > d/2, i.e.,
Z t
1
1
u ∈ Lt∞ H xs (I × Rd ),
u(t) = e 2 i(t−t0 )∆ u0 − iµ
e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]dτ,
t0
with the integral equation holding almost everywhere. Fixing two nearby times t, t0 ∈ I for
which the above integral equation holds, then we can bound
1
1
0
ku(t0 ) − u(t)kHxs ≤ k[e 2 i(t −t0 )∆ − e 2 i(t−t0 )∆ ]u0 kHxs
Z t0
Z t
1
1
0
+ e 2 i(t −τ)∆ [|u(τ)| p−1 u(τ)]dτ −
e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]dτ
t0
H s
t0
x
Z t0
1
1
1
0
≤ k[e 2 i(t −t0 )∆ − e 2 i(t−t0 )∆ ]u0 kHxs + e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]dτ
H s
t
x
Z t0
1 i(t0 −t)∆
1
− 1]
+ [e 2
e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]dτ
t0
H xs
= L + N1 + N2 .
By the dominated convergence theorem, the linear part L satisfies
Z
0
2
2 2
L2 =
(1 + |ξ|2 ) s ei(t −t0 )|ξ| − ei(t−t0 )|ξ| |û0 (ξ)|2 dξ → 0
Rd
as t0 → t, hence it is continuous in time. For the nonlinear part N1 , we use Lemma A.8, in
particular (A.18), along with the fact that the u(τ)’s are uniformly bounded in H s :
Z t0
N1 .
ku(τ)kHp xs dτ ≤ (t0 − t)kukLp∞ Hxs .
t
t
In particular, the right-hand side goes to 0 as t0 → t. Similarly, for the remaining term N2 ,
letting F denote the Fourier transform in the spatial variables, then
Z
2 21
0
2 Z t0
1
2
i(t−τ)∆
p−1
2
s
i(t
−t)|ξ|
[|u(τ)| u(τ)]}dτ dξ
N2 = (1 + |ξ| ) e
− 1 F {e 2
t0
Rd
Z t0 "Z
0
2 2 # 21
1
2
≤
(1 + |ξ|2 ) s ei(t −t)|ξ| − 1 F {e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]} dξ dτ
t0
0
Rd
≤ (t − t0 )
1
2
"Z
t0
Z
# 12
2 0
2
1
i(t −t)|ξ|2
i(t−τ)∆
p−1
2
− 1 F {e
(1 + |ξ| ) e
[|u(τ)| u(τ)]} dξdτ .
2 s
t0
Rd
By Lemma A.8, there is some constant C > 0, independent of τ, such that
Z
1
(1 + |ξ|2 ) s F {e 2 i(t−τ)∆ [|u(τ)| p−1 u(τ)]}dξ . kukLp∞ Hxs < C.
Rd
t
Thus, by the dominated convergence theorem, it follows that N2 → 0 as t0 → t.
Consequently, u can be considered (by replacing a subset of measure zero) as a continuous function into H s (Rd ), hence u is a strong H s -solution.
SOLUTIONS
47
3.13. Let J be any time interval containing t0 , and let u, v be two strong solutions to (3.1)
on J with the same initial data at t0 . Define the subset
A = {t ∈ J | u(t) = v(t)}.
Since both u and v are continuous with respect to t, then A is closed. Furthermore, if s ∈ A,
then by the local uniqueness assumption, there is an open interval I containing s such that
u(t) = v(t) for any t ∈ I. As a result, I ⊆ A, and it follows that A is open. Since A is
open, closed, and nonempty (since t0 ∈ A by assumption), it follows from connectedness
considerations that A = J. Thus, u and v coincide everywhere on J.
3.14. First, note that for any x ∈ Rd , we have, by definition and induction, the inequality
|∇ j hxik | .d,k hxik− j ,
for any nonnegative integers 0 ≤ j ≤ k, where ∇ denotes the spatial gradient on Rd . As a
result, we can apply the above in conjunction with the Leibniz rule in order to obtain
k f kHxk,k (Rd ) =
k
X
khxi j f kHxk− j (Rd ) .
j=0
X
X
k∇b (hxia f )kL2x (Rd ) .
a+b≤k
khxia ∇b f kL2x (Rd ) .
a+b≤k
Similarly, again by the above pointwise inequality and the Leibniz rule, we have
X
X
khxia ∇b f kL2x (Rd ) .
k∇b (hxia f )kL2x (Rd ) . k f kHxk,k (Rd ) .
a+b≤k
a+b≤k
Using the above equivalent formulation of the H k,k -norm and Hölder’s inequality yields
X
k f gkHxk,k (Rd ) .
khxia ∇b f ∇c gkL2x (Rd )
a+b+c+p=k
.
X
khxia ∇b f k
a+b+c+p=k
2k
L xa+b (Rd )
k∇c f k
2k
c+p
Lx
(Rd )
.
Given a, b, c, p as in the terms on the right-hand side above, since k > d/2, we have
!
a b
d
c p
d
1− −
< k − a − b,
1− −
< k − c − p.
2
k k
2
k k
Thus, the Gagliardo-Nirenberg inequality (Proposition A.3) yields
X
k f gkHxk,k (Rd ) .
khxia ∇b f kHxk−a−b (Rd ) k∇c f kHxk−c−p (Rd ) .
a+b+c+p=k
Finally, returning to our pointwise inequality, we have, as desired,
X
X
k f gkHxk,k (Rd ) .
khxia ∇b f kL2x (Rd )
k∇c f kL2x (Rd ) . k f kHxk,k (Rd ) kgkHxk,k (Rd ) .
a+b≤k
c+p≤k
Appendix A: Tools from Harmonic Analysis
A.15. Correction: For the first inequality, we need an extra condition, e.g., u having zero
mean. Otherwise, we can consider a constant function u ≡ c , 0, for which we have
kukL∞ (I) = c , 0,
kuk1/2
k∂ uk1/2 = 0.
L2 (I) t L2 (I)
48
ARICK SHAO
With the extra mean-free assumption for u, we can conclude via the intermediate value
theorem that u(t0 ) = 0 for some t0 ∈ I. 35 By the fundamental theorem of calculus,
Z t
Z t
2
2
|u(t)| =
∂t |u| .
|u||∂t u| . kukL2 (I) k∂t ukL2 (I) ,
t ∈ I.
t0
t0
This proves the Gagliardo-Nirenberg inequality.
Next, suppose I = [a, b], and let u denote the mean of u on I. As a first step, we assume
that u = 0. Integrating by parts, then we obtain
Z b
Z b"
Z t #
2
|u| =
u(t) · ∂t
u dt
a
a
a
= u(b) ·
Z
b
u−
a
=−
Z
b
b
Z
"
Z t #
∂t u(t) ·
u dt
a
a
"
Z t #
∂t u(t) ·
u dt.
a
a
Applying Hölder’s inequality yields
kuk2L2 (I)
Z
≤ k∂t ukL2 (I)
b
Z
b
a
Z
t
a
≤ k∂t ukL2 (I)
a
!2 21
u dt
! 21
|I|kuk2L2 (I)
≤ |I|k∂t ukL2 (I) kukL2 (I) ,
which implies the Poincaré inequality in the case u = 0.
Finally, for general u, since u − u has zero mean, then
ku − ukL2 (I) ≤ |I|k∂t (u − u)kL2 (I) = |I|k∂t ukL2 (I) .
References
1. B. Birnir, C. Kenig, G. Ponce, N. Svanstedt, and L. Vega, On the ill-posedness of the IVP for the generalized
Korteweg-de Vries and nonlinear Schrödinger equations, J. London Math. Soc. 53 (1996), 551–559.
2. J. Colliander, M. Keel, G. Staffilani, H. Takaoka, and T. Tao, Global well-posedness and scattering in the
energy space for the critical nonlinear Schrödinger equation in R3 , Annals of Math. 167 (2007), 767–865.
3. W. Rudin, Principles of mathematical analysis, McGraw-Hill, Inc., 1976.
4. E. Stein, Harmonic analysis: Real variable methods, orthogonality, and oscillatory integrals, Princeton University Press, 1993.
5. T. Tao, Nonlinear dispersive equations: Local and global analysis, American Mathematical Society, 2006.
35By a standard limiting argument, we can always assume that u is smooth on the interior of I.
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